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Class 9 ବୀଜଗଣିତ
ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(c)

ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(c) – Book Q A Class 9 ବୀଜଗଣିତ

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📝 ଅନୁଶୀଳନୀ - 3 ©

1. ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ ଠିକ୍ ଉତ୍ତରଟିକୁ ବାଛି ଲେଖ ।
(i)

❓ Question: x2−3x+2x^2 - 3x + 2 ର ଉତ୍ପାଦକ ଦ୍ୱୟ

(a) (x−2)(x - 2) ଓ (x+1)(x + 1)

(b) (x+2)(x + 2) ଓ (x−1)(x - 1)

© (x−2)(x - 2) ଓ (x−1)(x - 1)

(d) (x+2)(x + 2) ଓ (x+1)(x + 1)

✅ Answer: © (x−2)(x - 2) ଓ (x−1)(x - 1)

(କାରଣ: x2−3x+2=x2−2x−x+2=x(x−2)−1(x−2)=(x−2)(x−1)x^2 - 3x + 2 = x^2 - 2x - x + 2 = x(x - 2) - 1(x - 2) = (x - 2)(x - 1))

(ii)

❓ Question: ଏକ ଦ୍ୱିଘାତୀ ପଲିନୋମିଆଲର ଉତ୍ପାଦକ ଦ୍ୱୟ (x−1)(x - 1) ଓ (x−3)(x - 3) ହେଲେ ପଲିନୋମିଆଲଟି

(a) x2−4x−3x^2 - 4x - 3

(b) x2−4x+3x^2 - 4x + 3

© x2+4x−3x^2 + 4x - 3

(d) x2+4x+3x^2 + 4x + 3

✅ Answer: (b) x2−4x+3x^2 - 4x + 3

(କାରଣ: (x−1)(x−3)=x2−3x−x+3=x2−4x+3(x - 1)(x - 3) = x^2 - 3x - x + 3 = x^2 - 4x + 3)

(iii)

❓ Question: x4−y4x^4 - y^4 ର ଠିକ୍ ଉତ୍ପାଦକୀକରଣ ବାଛ ।

(a) (x2+y2)(x+y)(x−y)(x^2 + y^2)(x + y)(x - y)

(b) (x2−y2)(x−y)(x+y)(x^2 - y^2)(x - y)(x + y)

© (x2+y2)(x+y)2(x^2 + y^2)(x + y)^2

(d) (x2+y2)(x−y)2(x^2 + y^2)(x - y)^2

✅ Answer: (a) (x2+y2)(x+y)(x−y)(x^2 + y^2)(x + y)(x - y)

(କାରଣ: x4−y4=(x2)2−(y2)2=(x2−y2)(x2+y2)=(x−y)(x+y)(x2+y2)x^4 - y^4 = (x^2)^2 - (y^2)^2 = (x^2 - y^2)(x^2 + y^2) = (x - y)(x + y)(x^2 + y^2))

(iv)

❓ Question: 8a3−b3−12a2b+6ab28a^3 - b^3 - 12a^2b + 6ab^2 ର ଉତ୍ପାଦକ ଗୁଡିକ

(a) (2a−b),(2a+b),(2a+b)(2a - b), (2a + b), (2a + b)

(b) (2a+b)(2a+b)(2a+b)(2a + b)(2a + b)(2a + b)

© (2a−b),(2a−b),(2a+b)(2a - b), (2a - b), (2a + b)

(d) (2a−b),(2a−b),(2a−b)(2a - b), (2a - b), (2a - b)

✅ Answer: (d) (2a−b),(2a−b),(2a−b)(2a - b), (2a - b), (2a - b)

(କାରଣ: ଏହା (x−y)3=x3−y3−3x2y+3xy2(x - y)^3 = x^3 - y^3 - 3x^2y + 3xy^2 ସୂତ୍ର ଅଟେ, ଯେଉଁଠାରେ x=2ax = 2a ଏବଂ y=by = b । ଅର୍ଥାତ୍ (2a−b)3(2a - b)^3)

(v)

❓ Question: 625+25x4+x8625 + 25x^4 + x^8 ର ଉତ୍ପାଦକ ଗୁଡିକ ନିମ୍ନସ୍ଥ କେଉଁଟି 625+25x4+x8625 + 25x^4 + x^8 ର ଗୁଣଫଳ ସହ ସମାନ ।

(a) (25+5x2+x4),(25−5x2+x4)(25 + 5x^2 + x^4), (25 - 5x^2 + x^4)

(b) (25+5x2+x4),(25+5x2−x4)(25 + 5x^2 + x^4), (25 + 5x^2 - x^4)

© (25+5x4+x4),(25−5x4+x4)(25 + 5x^4 + x^4), (25 - 5x^4 + x^4)

(d) (25−5x4+x4),(25+5x4−x4)(25 - 5x^4 + x^4), (25 + 5x^4 - x^4)

✅ Answer: (a) (25+5x2+x4),(25−5x2+x4)(25 + 5x^2 + x^4), (25 - 5x^2 + x^4)

(କାରଣ: ଆସନ୍ତୁ ଗୁଣନ କରିବା - (x4+25+5x2)(x4+25−5x2)=(x4+25)2−(5x2)2=x8+50x4+625−25x4=x8+25x4+625(x^4 + 25 + 5x^2)(x^4 + 25 - 5x^2) = (x^4 + 25)^2 - (5x^2)^2 = x^8 + 50x^4 + 625 - 25x^4 = x^8 + 25x^4 + 625)
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(vi)

❓ Question: 1−a3+b3+3ab1 - a^3 + b^3 + 3ab ର ଗୋଟିଏ ଉତ୍ପାଦକ

(a) (1−a+b)(1 - a + b)

(b) (1−a−b)(1 - a - b)

© (1+a+b)(1 + a + b)

(d) (1+a−b)(1 + a - b)

✅ Answer: (a) (1−a+b)(1 - a + b)

(କାରଣ: ରାଶିଟିକୁ ଆମେ ଏହିପରି ଲେଖିପାରିବା 13+(−a)3+b3−3(1)(−a)(b)1^3 + (-a)^3 + b^3 - 3(1)(-a)(b) । ସୂତ୍ର x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) ଅନୁଯାୟୀ, ଗୋଟିଏ ଉତ୍ପାଦକ ହେବ (x+y+z)=(1−a+b)(x + y + z) = (1 - a + b) )
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(vii)

❓ Question: (2x−3y)3+(3y−4z)3+(4z−2x)3(2x - 3y)^3 + (3y - 4z)^3 + (4z - 2x)^3 ର ଉତ୍ପାଦକ ଗୁଡିକ ହେଲେ

(a) 6(2x−3y)(3y−4z)(2z−x)6(2x - 3y)(3y - 4z)(2z - x)

(b) 3(2x−3y)(3y−4z)(2z−x)3(2x - 3y)(3y - 4z)(2z - x)

© 60xyz60xyz

(d) ଏଥି ମଧ୍ୟରୁ କୌଣସିଟି ନୁହେଁ ।

✅ Answer: (a) 6(2x−3y)(3y−4z)(2z−x)6(2x - 3y)(3y - 4z)(2z - x)

(କାରଣ: ଯଦି A+B+C=0A + B + C = 0 ହୁଏ, ତେବେ A3+B3+C3=3ABCA^3 + B^3 + C^3 = 3ABC । ଏଠାରେ (2x−3y)+(3y−4z)+(4z−2x)=0(2x - 3y) + (3y - 4z) + (4z - 2x) = 0 ଅଟେ । ତେଣୁ ମାନ ହେବ: 3(2x−3y)(3y−4z)(4z−2x)3(2x - 3y)(3y - 4z)(4z - 2x) । ଏହାକୁ ଆହୁରି ସରଳ କଲେ (4z−2x)(4z - 2x) ରୁ 22 କମନ୍ ଆସିବ: 3(2x−3y)(3y−4z)×2(2z−x)=6(2x−3y)(3y−4z)(2z−x)3(2x - 3y)(3y - 4z) \times 2(2z - x) = 6(2x - 3y)(3y - 4z)(2z - x) )

(viii)

❓ Question: (28)3+(−15)3+(−13)3(28)^3 + (-15)^3 + (-13)^3 ର ସରଳୀକୃତ ମାନ

(a) 81908190

(b) 1638016380

© 2457024570

(d) 40954095

✅ Answer: (b) 1638016380

(କାରଣ: ଏଠାରେ a+b+c=28+(−15)+(−13)=0a + b + c = 28 + (-15) + (-13) = 0 । ତେଣୁ ସୂତ୍ର ଅନୁଯାୟୀ a3+b3+c3=3abca^3 + b^3 + c^3 = 3abc ହେବ । ∴3×28×(−15)×(−13)=3×28×195=16380\therefore 3 \times 28 \times (-15) \times (-13) = 3 \times 28 \times 195 = 16380 )

(ix)

❓ Question: (a−b)3+(b−c)3+(c−a)3(a - b)^3 + (b - c)^3 + (c - a)^3 ର ମାନ

(a) 3abc3abc

(b) 3a3b3c33a^3b^3c^3

© 3(a−b)(b−c)(c−a)3(a - b)(b - c)(c - a)

(d) {a−(b+c)}3\{a - (b + c)\}^3

✅ Answer: © 3(a−b)(b−c)(c−a)3(a - b)(b - c)(c - a)

(କାରଣ: ଏଠାରେ ପଦଗୁଡ଼ିକର ଯୋଗଫଳ (a−b)+(b−c)+(c−a)=0(a - b) + (b - c) + (c - a) = 0 । ତେଣୁ ସୂତ୍ର X3+Y3+Z3=3XYZX^3 + Y^3 + Z^3 = 3XYZ ଅନୁଯାୟୀ ମାନ ହେବ 3(a−b)(b−c)(c−a)3(a - b)(b - c)(c - a) )

(x)

❓ Question: 2x2−x−12x^2 - x - 1 ର ଗୋଟିଏ ଉତ୍ପାଦକ

(a) 2x−12x - 1

(b) x+1x + 1

© x−1x - 1

(d) x+2x + 2

✅ Answer: © x−1x - 1

(କାରଣ: ମଧ୍ୟପଦକୁ ଭାଙ୍ଗିଲେ, 2x2−x−1=2x2−2x+x−1=2x(x−1)+1(x−1)=(x−1)(2x+1)2x^2 - x - 1 = 2x^2 - 2x + x - 1 = 2x(x - 1) + 1(x - 1) = (x - 1)(2x + 1) । ତେଣୁ ଏହାର ଗୋଟିଏ ଉତ୍ପାଦକ ହେଉଛି x−1x - 1 )

Question -2 💡 ଉତ୍ପାଦକ ବିଶ୍ଳେଷଣର କର :

ax2+bx+cax^2 + bx + c ସମୀକରଣରେ, ଏପରି ଦୁଇଟି ସଂଖ୍ୟା ଖୋଜନ୍ତୁ ଯାହାର ଗୁଣଫଳ a×ca \times c ହେଉଥିବ ଏବଂ ଯୋଗଫଳ ମଧ୍ୟପଦ bb ସହ ସମାନ ହେଉଥିବ।

(i)

❓ ପ୍ରଶ୍ନ: 2x2−x−12x^2 - x - 1

✅ ଉତ୍ତର:

(ଟ୍ରିକ୍: ଗୁଣଫଳ 2×(−1)=−22 \times (-1) = -2, ଯୋଗଫଳ −1-1 । ତେଣୁ ସଂଖ୍ୟା ଦ୍ୱୟ ହେଲା −2-2 ଓ 11)

=2x2−2x+x−1= 2x^2 - 2x + x - 1

=2x(x−1)+1(x−1)= 2x(x - 1) + 1(x - 1)

=(x−1)(2x+1)= (x - 1)(2x + 1)

(ii)

❓ ପ୍ରଶ୍ନ: 2x2−3x+12x^2 - 3x + 1

✅ ଉତ୍ତର:

(ଟ୍ରିକ୍: ଗୁଣଫଳ 2×1=22 \times 1 = 2, ଯୋଗଫଳ −3-3 । ତେଣୁ ସଂଖ୍ୟା ଦ୍ୱୟ ହେଲା −2-2 ଓ −1-1)

=2x2−2x−x+1= 2x^2 - 2x - x + 1

=2x(x−1)−1(x−1)= 2x(x - 1) - 1(x - 1)

=(x−1)(2x−1)= (x - 1)(2x - 1)

(iii)

❓ ପ୍ରଶ୍ନ: 5x2−x−45x^2 - x - 4

✅ ଉତ୍ତର:

(ଟ୍ରିକ୍: ଗୁଣଫଳ 5×(−4)=−205 \times (-4) = -20, ଯୋଗଫଳ −1-1 । ତେଣୁ ସଂଖ୍ୟା ଦ୍ୱୟ ହେଲା −5-5 ଓ 44)

=5x2−5x+4x−4= 5x^2 - 5x + 4x - 4

=5x(x−1)+4(x−1)= 5x(x - 1) + 4(x - 1)

=(x−1)(5x+4)= (x - 1)(5x + 4)

(iv)

❓ ପ୍ରଶ୍ନ: 4x2−5x−64x^2 - 5x - 6

✅ ଉତ୍ତର:

(ଟ୍ରିକ୍: ଗୁଣଫଳ 4×(−6)=−244 \times (-6) = -24, ଯୋଗଫଳ −5-5 । ତେଣୁ ସଂଖ୍ୟା ଦ୍ୱୟ ହେଲା −8-8 ଓ 33)

=4x2−8x+3x−6= 4x^2 - 8x + 3x - 6

=4x(x−2)+3(x−2)= 4x(x - 2) + 3(x - 2)

=(x−2)(4x+3)= (x - 2)(4x + 3)

(v)

❓ ପ୍ରଶ୍ନ: 3x2+11x+63x^2 + 11x + 6

✅ ଉତ୍ତର:

(ଟ୍ରିକ୍: ଗୁଣଫଳ 3×6=183 \times 6 = 18, ଯୋଗଫଳ 1111 । ତେଣୁ ସଂଖ୍ୟା ଦ୍ୱୟ ହେଲା 99 ଓ 22)

=3x2+9x+2x+6= 3x^2 + 9x + 2x + 6

=3x(x+3)+2(x+3)= 3x(x + 3) + 2(x + 3)

=(x+3)(3x+2)= (x + 3)(3x + 2)

(vi)

❓ ପ୍ରଶ୍ନ: 7x2+x−67x^2 + x - 6

✅ ଉତ୍ତର:

(ଟ୍ରିକ୍: ଗୁଣଫଳ 7×(−6)=−427 \times (-6) = -42, ଯୋଗଫଳ 11 । ତେଣୁ ସଂଖ୍ୟା ଦ୍ୱୟ ହେଲା 77 ଓ −6-6)

=7x2+7x−6x−6= 7x^2 + 7x - 6x - 6

=7x(x+1)−6(x+1)= 7x(x + 1) - 6(x + 1)

=(x+1)(7x−6)= (x + 1)(7x - 6)

(vii)

❓ ପ୍ରଶ୍ନ: 2x2+5x−72x^2 + 5x - 7

✅ ଉତ୍ତର:

(ଟ୍ରିକ୍: ଗୁଣଫଳ 2×(−7)=−142 \times (-7) = -14, ଯୋଗଫଳ 55 । ତେଣୁ ସଂଖ୍ୟା ଦ୍ୱୟ ହେଲା 77 ଓ −2-2)

=2x2−2x+7x−7= 2x^2 - 2x + 7x - 7

=2x(x−1)+7(x−1)= 2x(x - 1) + 7(x - 1)

=(x−1)(2x+7)= (x - 1)(2x + 7)

(viii)

❓ ପ୍ରଶ୍ନ: 4x2−5x+14x^2 - 5x + 1

✅ ଉତ୍ତର:

(ଟ୍ରିକ୍: ଗୁଣଫଳ 4×1=44 \times 1 = 4, ଯୋଗଫଳ −5-5 । ତେଣୁ ସଂଖ୍ୟା ଦ୍ୱୟ ହେଲା −4-4 ଓ −1-1)

=4x2−4x−x+1= 4x^2 - 4x - x + 1

=4x(x−1)−1(x−1)= 4x(x - 1) - 1(x - 1)

=(x−1)(4x−1)= (x - 1)(4x - 1)

(ix)

❓ ପ୍ରଶ୍ନ: 4x2−3x−74x^2 - 3x - 7

✅ ଉତ୍ତର:

(ଟ୍ରିକ୍: ଗୁଣଫଳ 4×(−7)=−284 \times (-7) = -28, ଯୋଗଫଳ −3-3 । ତେଣୁ ସଂଖ୍ୟା ଦ୍ୱୟ ହେଲା −7-7 ଓ 44)

=4x2+4x−7x−7= 4x^2 + 4x - 7x - 7

=4x(x+1)−7(x+1)= 4x(x + 1) - 7(x + 1)

=(x+1)(4x−7)= (x + 1)(4x - 7)
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🧩 3. ଉତ୍ପାଦକରେ ବିଶ୍ଳେଷଣ କର ।

(i)

❓ ପ୍ରଶ୍ନ: 25a4−16b225a^4 - 16b^2

✅ ଉତ୍ତର:

=(5a2)2−(4b)2= (5a^2)^2 - (4b)^2

(ସୂତ୍ର x2−y2=(x+y)(x−y)x^2 - y^2 = (x + y)(x - y) ପ୍ରୟୋଗ କରି)

=(5a2+4b)(5a2−4b)= (5a^2 + 4b)(5a^2 - 4b)

(ii)

❓ ପ୍ରଶ୍ନ: 9−64p2q29 - 64p^2q^2

✅ ଉତ୍ତର:

=32−(8pq)2= 3^2 - (8pq)^2

=(3+8pq)(3−8pq)= (3 + 8pq)(3 - 8pq)

(iii)

❓ ପ୍ରଶ୍ନ: 8x3+27y38x^3 + 27y^3

✅ ଉତ୍ତର:

=(2x)3+(3y)3= (2x)^3 + (3y)^3

(ସୂତ୍ର a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2 - ab + b^2) ପ୍ରୟୋଗ କରି)

=(2x+3y)((2x)2−(2x)(3y)+(3y)2)= (2x + 3y)((2x)^2 - (2x)(3y) + (3y)^2)

=(2x+3y)(4x2−6xy+9y2)= (2x + 3y)(4x^2 - 6xy + 9y^2)

(iv)

❓ ପ୍ରଶ୍ନ: 8x3−27y38x^3 - 27y^3

✅ ଉତ୍ତର:

=(2x)3−(3y)3= (2x)^3 - (3y)^3

(ସୂତ୍ର a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2) ପ୍ରୟୋଗ କରି)

=(2x−3y)((2x)2+(2x)(3y)+(3y)2)= (2x - 3y)((2x)^2 + (2x)(3y) + (3y)^2)

=(2x−3y)(4x2+6xy+9y2)= (2x - 3y)(4x^2 + 6xy + 9y^2)

(v)

❓ ପ୍ରଶ୍ନ: (a+b)2−9(a+b)^2 - 9

✅ ଉତ୍ତର:

=(a+b)2−32= (a+b)^2 - 3^2

=(a+b+3)(a+b−3)= (a + b + 3)(a + b - 3)

(vi)

❓ ପ୍ରଶ୍ନ: (2a+5)2−16(2a+5)^2 - 16

✅ ଉତ୍ତର:

=(2a+5)2−42= (2a+5)^2 - 4^2

=(2a+5+4)(2a+5−4)= (2a + 5 + 4)(2a + 5 - 4)

=(2a+9)(2a+1)= (2a + 9)(2a + 1)

(vii)

❓ ପ୍ରଶ୍ନ: (x+2y)2−(x−y)2(x+2y)^2 - (x-y)^2

✅ ଉତ୍ତର:

={(x+2y)+(x−y)}{(x+2y)−(x−y)}= \{(x+2y) + (x-y)\}\{(x+2y) - (x-y)\}

=(x+2y+x−y)(x+2y−x+y)= (x + 2y + x - y)(x + 2y - x + y)

=(2x+y)(3y)= (2x + y)(3y)

=3y(2x+y)= 3y(2x + y)

(viii)

❓ ପ୍ରଶ୍ନ: 4(a+2p)2−9(2a−p)24(a+2p)^2 - 9(2a-p)^2

✅ ଉତ୍ତର:

=[2(a+2p)]2−[3(2a−p)]2= [2(a+2p)]^2 - [3(2a-p)]^2

=(2a+4p)2−(6a−3p)2= (2a + 4p)^2 - (6a - 3p)^2

={(2a+4p)+(6a−3p)}{(2a+4p)−(6a−3p)}= \{(2a + 4p) + (6a - 3p)\}\{(2a + 4p) - (6a - 3p)\}

=(2a+4p+6a−3p)(2a+4p−6a+3p)= (2a + 4p + 6a - 3p)(2a + 4p - 6a + 3p)

=(8a+p)(7p−4a)= (8a + p)(7p - 4a)

(ix)

❓ ପ୍ରଶ୍ନ: 75(2a−b+1)2−12(a+b)275(2a-b+1)^2 - 12(a+b)^2

✅ ଉତ୍ତର:

33 କୁ ସାଧାରଣ ରାଶି ଭାବେ ନେଲେ:

=3[25(2a−b+1)2−4(a+b)2]= 3[25(2a-b+1)^2 - 4(a+b)^2]

=3[{5(2a−b+1)}2−{2(a+b)}2]= 3[\{5(2a-b+1)\}^2 - \{2(a+b)\}^2]

=3[(10a−5b+5)2−(2a+2b)2]= 3[(10a - 5b + 5)^2 - (2a + 2b)^2]

=3{(10a−5b+5)+(2a+2b)}{(10a−5b+5)−(2a+2b)}= 3\{(10a - 5b + 5) + (2a + 2b)\}\{(10a - 5b + 5) - (2a + 2b)\}

=3(10a−5b+5+2a+2b)(10a−5b+5−2a−2b)= 3(10a - 5b + 5 + 2a + 2b)(10a - 5b + 5 - 2a - 2b)

=3(12a−3b+5)(8a−7b+5)= 3(12a - 3b + 5)(8a - 7b + 5)

(x)

❓ ପ୍ରଶ୍ନ: (a+b)3−8c3(a+b)^3 - 8c^3

✅ ଉତ୍ତର:

=(a+b)3−(2c)3= (a+b)^3 - (2c)^3

={(a+b)−2c}{(a+b)2+(a+b)(2c)+(2c)2}= \{(a+b) - 2c\}\{(a+b)^2 + (a+b)(2c) + (2c)^2\}

=(a+b−2c)(a2+2ab+b2+2ac+2bc+4c2)= (a + b - 2c)(a^2 + 2ab + b^2 + 2ac + 2bc + 4c^2)

(xi)

❓ ପ୍ରଶ୍ନ: p4−27pq6p^4 - 27pq^6

✅ ଉତ୍ତର:

pp କୁ ସାଧାରଣ ରାଶି ଭାବେ ନେଲେ:

=p(p3−27q6)= p(p^3 - 27q^6)

=p{p3−(3q2)3}= p\{p^3 - (3q^2)^3\}

=p(p−3q2){p2+p(3q2)+(3q2)2}= p(p - 3q^2)\{p^2 + p(3q^2) + (3q^2)^2\}

=p(p−3q2)(p2+3pq2+9q4)= p(p - 3q^2)(p^2 + 3pq^2 + 9q^4)

(xii)

❓ ପ୍ରଶ୍ନ: 1−(a+2)31 - (a+2)^3

✅ ଉତ୍ତର:

=13−(a+2)3= 1^3 - (a+2)^3

={1−(a+2)}{12+1(a+2)+(a+2)2}= \{1 - (a+2)\}\{1^2 + 1(a+2) + (a+2)^2\}

=(1−a−2)(1+a+2+a2+4a+4)= (1 - a - 2)(1 + a + 2 + a^2 + 4a + 4)

=(−a−1)(a2+5a+7)= (-a - 1)(a^2 + 5a + 7)

=−(a+1)(a2+5a+7)= -(a + 1)(a^2 + 5a + 7)

(xiii)

❓ ପ୍ରଶ୍ନ: 8−(2x−3)38 - (2x-3)^3

✅ ଉତ୍ତର:

=23−(2x−3)3= 2^3 - (2x-3)^3

={2−(2x−3)}{22+2(2x−3)+(2x−3)2}= \{2 - (2x-3)\}\{2^2 + 2(2x-3) + (2x-3)^2\}

=(2−2x+3)(4+4x−6+4x2−12x+9)= (2 - 2x + 3)(4 + 4x - 6 + 4x^2 - 12x + 9)

=(5−2x)(4x2−8x+7)= (5 - 2x)(4x^2 - 8x + 7)

(xiv)

❓ ପ୍ରଶ୍ନ: 320p6q−5p2q7320p^6q - 5p^2q^7

✅ ଉତ୍ତର:

5p2q5p^2q କୁ ସାଧାରଣ ରାଶି ଭାବେ ନେଲେ:

=5p2q(64p4−q6)= 5p^2q(64p^4 - q^6)

=5p2q{(8p2)2−(q3)2}= 5p^2q\{(8p^2)^2 - (q^3)^2\}

=5p2q(8p2+q3)(8p2−q3)= 5p^2q(8p^2 + q^3)(8p^2 - q^3)

(xv)

❓ ପ୍ରଶ୍ନ: 1+(a+2)31 + (a+2)^3

✅ ଉତ୍ତର:

=13+(a+2)3= 1^3 + (a+2)^3

={1+(a+2)}{12−1(a+2)+(a+2)2}= \{1 + (a+2)\}\{1^2 - 1(a+2) + (a+2)^2\}

=(1+a+2)(1−a−2+a2+4a+4)= (1 + a + 2)(1 - a - 2 + a^2 + 4a + 4)

=(a+3)(a2+3a+3)= (a + 3)(a^2 + 3a + 3)

(xvi)

❓ ପ୍ରଶ୍ନ: 8+(2x−3)38 + (2x-3)^3

✅ ଉତ୍ତର:

=23+(2x−3)3= 2^3 + (2x-3)^3

={2+(2x−3)}{22−2(2x−3)+(2x−3)2}= \{2 + (2x-3)\}\{2^2 - 2(2x-3) + (2x-3)^2\}

=(2+2x−3)(4−4x+6+4x2−12x+9)= (2 + 2x - 3)(4 - 4x + 6 + 4x^2 - 12x + 9)

=(2x−1)(4x2−16x+19)= (2x - 1)(4x^2 - 16x + 19)

(xvii)

❓ ପ୍ରଶ୍ନ: a3+6a2b+12ab2+8b3a^3 + 6a^2b + 12ab^2 + 8b^3

✅ ଉତ୍ତର:

=a3+3(a2)(2b)+3(a)(2b)2+(2b)3= a^3 + 3(a^2)(2b) + 3(a)(2b)^2 + (2b)^3

(ସୂତ୍ର x3+3x2y+3xy2+y3=(x+y)3x^3 + 3x^2y + 3xy^2 + y^3 = (x + y)^3 ପ୍ରୟୋଗ କରି)

=(a+2b)3= (a + 2b)^3

(xviii)

❓ ପ୍ରଶ୍ନ: a3+9a2+27a+27a^3 + 9a^2 + 27a + 27

✅ ଉତ୍ତର:

=a3+3(a2)(3)+3(a)(32)+33= a^3 + 3(a^2)(3) + 3(a)(3^2) + 3^3

=(a+3)3= (a + 3)^3

(xix)

❓ ପ୍ରଶ୍ନ: 8−36p+54p2−27p38 - 36p + 54p^2 - 27p^3

✅ ଉତ୍ତର:

=23−3(22)(3p)+3(2)(3p)2−(3p)3= 2^3 - 3(2^2)(3p) + 3(2)(3p)^2 - (3p)^3

(ସୂତ୍ର x3−3x2y+3xy2−y3=(x−y)3x^3 - 3x^2y + 3xy^2 - y^3 = (x - y)^3 ପ୍ରୟୋଗ କରି)

=(2−3p)3= (2 - 3p)^3

(xx)

❓ ପ୍ରଶ୍ନ: (b−q)3−(c−q)3−3(b−c)(b−q)(c−q)(b-q)^3 - (c-q)^3 - 3(b-c)(b-q)(c-q)

✅ ଉତ୍ତର:

ମନେକର x=(b−q)x = (b-q) ଏବଂ y=(c−q)y = (c-q) ।

ଏଠାରେ ଲକ୍ଷ୍ୟ କରନ୍ତୁ ଯେ x−y=(b−q)−(c−q)=b−q−c+q=b−cx - y = (b-q) - (c-q) = b - q - c + q = b - c ।

ଏହି ମାନଗୁଡ଼ିକୁ ରାଶିରେ ପ୍ରୟୋଗ କଲେ:

=x3−y3−3(x−y)xy= x^3 - y^3 - 3(x-y)xy

(ଆମେ ଜାଣୁ ଯେ ସୂତ୍ର x3−y3−3xy(x−y)=(x−y)3x^3 - y^3 - 3xy(x - y) = (x - y)^3)

=(x−y)3= (x - y)^3

=(b−c)3= (b - c)^3

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🧩 4. ଉତ୍ପାଦକରେ ବିଶ୍ଳେଷଣ କର ।

(i)

❓ ପ୍ରଶ୍ନ: a4+a2+1a^4 + a^2 + 1

✅ ଉତ୍ତର:

=(a2)2+2a2+1−a2= (a^2)^2 + 2a^2 + 1 - a^2

=(a2+1)2−a2= (a^2 + 1)^2 - a^2

=(a2+1+a)(a2+1−a)= (a^2 + 1 + a)(a^2 + 1 - a)

=(a2+a+1)(a2−a+1)= (a^2 + a + 1)(a^2 - a + 1)

(ii)

❓ ପ୍ରଶ୍ନ: a4b4+a2b2+1a^4b^4 + a^2b^2 + 1

✅ ଉତ୍ତର:

=(a2b2)2+2a2b2+1−a2b2= (a^2b^2)^2 + 2a^2b^2 + 1 - a^2b^2

=(a2b2+1)2−(ab)2= (a^2b^2 + 1)^2 - (ab)^2

=(a2b2+1+ab)(a2b2+1−ab)= (a^2b^2 + 1 + ab)(a^2b^2 + 1 - ab)

=(a2b2+ab+1)(a2b2−ab+1)= (a^2b^2 + ab + 1)(a^2b^2 - ab + 1)

(iii)

❓ ପ୍ରଶ୍ନ: 16a4+36a2b2+81b416a^4 + 36a^2b^2 + 81b^4

✅ ଉତ୍ତର:

=(4a2)2+72a2b2+(9b2)2−36a2b2= (4a^2)^2 + 72a^2b^2 + (9b^2)^2 - 36a^2b^2

=(4a2+9b2)2−(6ab)2= (4a^2 + 9b^2)^2 - (6ab)^2

=(4a2+9b2+6ab)(4a2+9b2−6ab)= (4a^2 + 9b^2 + 6ab)(4a^2 + 9b^2 - 6ab)

=(4a2+6ab+9b2)(4a2−6ab+9b2)= (4a^2 + 6ab + 9b^2)(4a^2 - 6ab + 9b^2)

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(iv)

❓ ପ୍ରଶ୍ନ: a8+a4+1a^8 + a^4 + 1

✅ ଉତ୍ତର:

=(a4)2+2a4+1−a4= (a^4)^2 + 2a^4 + 1 - a^4

=(a4+1)2−(a2)2= (a^4 + 1)^2 - (a^2)^2

=(a4+a2+1)(a4−a2+1)= (a^4 + a^2 + 1)(a^4 - a^2 + 1)

(ବର୍ତ୍ତମାନ ପ୍ରଥମ ପଦକୁ ପୁଣିଥରେ ବିଶ୍ଳେଷଣ କଲେ)

=(a2+a+1)(a2−a+1)(a4−a2+1)= (a^2 + a + 1)(a^2 - a + 1)(a^4 - a^2 + 1)

(v)

❓ ପ୍ରଶ୍ନ: x4+4x^4 + 4

✅ ଉତ୍ତର:

=(x2)2+22+4x2−4x2= (x^2)^2 + 2^2 + 4x^2 - 4x^2

=(x2+2)2−(2x)2= (x^2 + 2)^2 - (2x)^2

=(x2+2+2x)(x2+2−2x)= (x^2 + 2 + 2x)(x^2 + 2 - 2x)

=(x2+2x+2)(x2−2x+2)= (x^2 + 2x + 2)(x^2 - 2x + 2)

(vi)

❓ ପ୍ରଶ୍ନ: 2a4+8b42a^4 + 8b^4

✅ ଉତ୍ତର:

22 କୁ ସାଧାରଣ ରାଶି ଭାବେ ନେଲେ:

=2(a4+4b4)= 2(a^4 + 4b^4)

=2{(a2)2+(2b2)2+4a2b2−4a2b2}= 2\{(a^2)^2 + (2b^2)^2 + 4a^2b^2 - 4a^2b^2\}

=2{(a2+2b2)2−(2ab)2}= 2\{(a^2 + 2b^2)^2 - (2ab)^2\}

=2(a2+2ab+2b2)(a2−2ab+2b2)= 2(a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2)

(vii)

❓ ପ୍ରଶ୍ନ: 36a4+9b436a^4 + 9b^4

✅ ଉତ୍ତର:

99 କୁ ସାଧାରଣ ରାଶି ଭାବେ ନେଲେ:

=9(4a4+b4)= 9(4a^4 + b^4)

=9{(2a2)2+(b2)2+4a2b2−4a2b2}= 9\{(2a^2)^2 + (b^2)^2 + 4a^2b^2 - 4a^2b^2\}

=9{(2a2+b2)2−(2ab)2}= 9\{(2a^2 + b^2)^2 - (2ab)^2\}

=9(2a2+2ab+b2)(2a2−2ab+b2)= 9(2a^2 + 2ab + b^2)(2a^2 - 2ab + b^2)

(viii)

❓ ପ୍ରଶ୍ନ: 4a4+7a2+164a^4 + 7a^2 + 16

✅ ଉତ୍ତର:

=(2a2)2+16a2+42−9a2= (2a^2)^2 + 16a^2 + 4^2 - 9a^2

=(2a2+4)2−(3a)2= (2a^2 + 4)^2 - (3a)^2

=(2a2+3a+4)(2a2−3a+4)= (2a^2 + 3a + 4)(2a^2 - 3a + 4)

(ix)

❓ ପ୍ରଶ୍ନ: a4+2a2b2+9b4a^4 + 2a^2b^2 + 9b^4

✅ ଉତ୍ତର:

=(a2)2+6a2b2+(3b2)2−4a2b2= (a^2)^2 + 6a^2b^2 + (3b^2)^2 - 4a^2b^2

=(a2+3b2)2−(2ab)2= (a^2 + 3b^2)^2 - (2ab)^2

=(a2+2ab+3b2)(a2−2ab+3b2)= (a^2 + 2ab + 3b^2)(a^2 - 2ab + 3b^2)

(x)

❓ ପ୍ରଶ୍ନ: a4−3a2+1a^4 - 3a^2 + 1

✅ ଉତ୍ତର:

=(a2)2−2a2+1−a2= (a^2)^2 - 2a^2 + 1 - a^2

=(a2−1)2−a2= (a^2 - 1)^2 - a^2

=(a2−1+a)(a2−1−a)= (a^2 - 1 + a)(a^2 - 1 - a)

=(a2+a−1)(a2−a−1)= (a^2 + a - 1)(a^2 - a - 1)

(xi)

❓ ପ୍ରଶ୍ନ: 25a4−19a2b2+9b425a^4 - 19a^2b^2 + 9b^4

✅ ଉତ୍ତର:

=(5a2)2+30a2b2+(3b2)2−49a2b2= (5a^2)^2 + 30a^2b^2 + (3b^2)^2 - 49a^2b^2

=(5a2+3b2)2−(7ab)2= (5a^2 + 3b^2)^2 - (7ab)^2

=(5a2+7ab+3b2)(5a2−7ab+3b2)= (5a^2 + 7ab + 3b^2)(5a^2 - 7ab + 3b^2)

(xii)

❓ ପ୍ରଶ୍ନ: 9x2+y2+6xy−4z29x^2 + y^2 + 6xy - 4z^2

✅ ଉତ୍ତର:

ସଜାଇ ଲେଖିଲେ:

=(3x)2+2(3x)(y)+(y)2−(2z)2= (3x)^2 + 2(3x)(y) + (y)^2 - (2z)^2

=(3x+y)2−(2z)2= (3x + y)^2 - (2z)^2

=(3x+y+2z)(3x+y−2z)= (3x + y + 2z)(3x + y - 2z)

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(xiii)

❓ ପ୍ରଶ୍ନ: 16−x2−24y+9y216 - x^2 - 24y + 9y^2

✅ ଉତ୍ତର:

ସଜାଇ ଲେଖିଲେ:

=16−24y+9y2−x2= 16 - 24y + 9y^2 - x^2

=(4)2−2(4)(3y)+(3y)2−x2= (4)^2 - 2(4)(3y) + (3y)^2 - x^2

=(4−3y)2−x2= (4 - 3y)^2 - x^2

=(4−3y+x)(4−3y−x)= (4 - 3y + x)(4 - 3y - x)

(xiv)

❓ ପ୍ରଶ୍ନ: (a2−b2)(x2−y2)−4abxy(a^2 - b^2)(x^2 - y^2) - 4abxy

✅ ଉତ୍ତର:

ଗୁଣନ କଲେ:

=a2x2−a2y2−b2x2+b2y2−2abxy−2abxy= a^2x^2 - a^2y^2 - b^2x^2 + b^2y^2 - 2abxy - 2abxy

ପଦଗୁଡିକୁ ଏକାଠି କଲେ:

=(a2x2−2abxy+b2y2)−(a2y2+2abxy+b2x2)= (a^2x^2 - 2abxy + b^2y^2) - (a^2y^2 + 2abxy + b^2x^2)

=(ax−by)2−(ay+bx)2= (ax - by)^2 - (ay + bx)^2

=(ax−by+ay+bx)(ax−by−ay−bx)= (ax - by + ay + bx)(ax - by - ay - bx)

(xv)

❓ ପ୍ରଶ୍ନ: (a2+b2)(x2−y2)−2ab(x2+y2)(a^2 + b^2)(x^2 - y^2) - 2ab(x^2 + y^2)

✅ ଉତ୍ତର:

ଗୁଣନ କଲେ:

=a2x2−a2y2+b2x2−b2y2−2abx2−2aby2= a^2x^2 - a^2y^2 + b^2x^2 - b^2y^2 - 2abx^2 - 2aby^2

ପଦଗୁଡିକୁ ଏକାଠି କଲେ:

=(a2x2−2abx2+b2x2)−(a2y2+2aby2+b2y2)= (a^2x^2 - 2abx^2 + b^2x^2) - (a^2y^2 + 2aby^2 + b^2y^2)

=x2(a2−2ab+b2)−y2(a2+2ab+b2)= x^2(a^2 - 2ab + b^2) - y^2(a^2 + 2ab + b^2)

=x2(a−b)2−y2(a+b)2= x^2(a - b)^2 - y^2(a + b)^2

={x(a−b)}2−{y(a+b)}2= \{x(a - b)\}^2 - \{y(a + b)\}^2

=(ax−bx+ay+by)(ax−bx−ay−by)= (ax - bx + ay + by)(ax - bx - ay - by)
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🧩 5. ଉତ୍ପାଦକରେ ବିଶ୍ଳେଷଣ କର ।

(i)

❓ ପ୍ରଶ୍ନ: a3+b3+x3−3abxa^3 + b^3 + x^3 - 3abx

✅ ଉତ୍ତର:

ସୂତ୍ର X3+Y3+Z3−3XYZ=(X+Y+Z)(X2+Y2+Z2−XY−YZ−ZX)X^3 + Y^3 + Z^3 - 3XYZ = (X + Y + Z)(X^2 + Y^2 + Z^2 - XY - YZ - ZX) ପ୍ରୟୋଗ କରିବା:

=(a+b+x)(a2+b2+x2−ab−bx−ax)= (a + b + x)(a^2 + b^2 + x^2 - ab - bx - ax)

(ii)

❓ ପ୍ରଶ୍ନ: 8a3+b3+c3−6abc8a^3 + b^3 + c^3 - 6abc

✅ ଉତ୍ତର:

=(2a)3+b3+c3−3(2a)(b)(c)= (2a)^3 + b^3 + c^3 - 3(2a)(b)(c)

=(2a+b+c)((2a)2+b2+c2−(2a)(b)−(b)(c)−(c)(2a))= (2a + b + c)((2a)^2 + b^2 + c^2 - (2a)(b) - (b)(c) - (c)(2a))

=(2a+b+c)(4a2+b2+c2−2ab−bc−2ca)= (2a + b + c)(4a^2 + b^2 + c^2 - 2ab - bc - 2ca)

(iii)

❓ ପ୍ରଶ୍ନ: a3+b3−8+6aba^3 + b^3 - 8 + 6ab

✅ ଉତ୍ତର:

=a3+b3+(−2)3−3(a)(b)(−2)= a^3 + b^3 + (-2)^3 - 3(a)(b)(-2)

=(a+b−2)(a2+b2+(−2)2−ab−(b)(−2)−(−2)(a))= (a + b - 2)(a^2 + b^2 + (-2)^2 - ab - (b)(-2) - (-2)(a))

=(a+b−2)(a2+b2+4−ab+2b+2a)= (a + b - 2)(a^2 + b^2 + 4 - ab + 2b + 2a)
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(iv)

❓ ପ୍ରଶ୍ନ: l3−27m3−n3−9lmnl^3 - 27m^3 - n^3 - 9lmn

✅ ଉତ୍ତର:

=l3+(−3m)3+(−n)3−3(l)(−3m)(−n)= l^3 + (-3m)^3 + (-n)^3 - 3(l)(-3m)(-n)

=(l−3m−n)(l2+(−3m)2+(−n)2−l(−3m)−(−3m)(−n)−(−n)(l))= (l - 3m - n)(l^2 + (-3m)^2 + (-n)^2 - l(-3m) - (-3m)(-n) - (-n)(l))

=(l−3m−n)(l2+9m2+n2+3lm−3mn+nl)= (l - 3m - n)(l^2 + 9m^2 + n^2 + 3lm - 3mn + nl)

(v)

❓ ପ୍ରଶ୍ନ: (a−b)3+(b−c)3+(c−a)3−3(a−b)(b−c)(c−a)(a - b)^3 + (b - c)^3 + (c - a)^3 - 3(a - b)(b - c)(c - a)

✅ ଉତ୍ତର:

ମନେକର X=a−bX = a - b, Y=b−cY = b - c, ଏବଂ Z=c−aZ = c - a ।

ଏଠାରେ ଲକ୍ଷ୍ୟ କରନ୍ତୁ ଯେ: X+Y+Z=a−b+b−c+c−a=0X + Y + Z = a - b + b - c + c - a = 0

ଆମେ ଜାଣୁ ଯେ ଯଦି X+Y+Z=0X + Y + Z = 0 ହୁଏ, ତେବେ X3+Y3+Z3−3XYZ=0X^3 + Y^3 + Z^3 - 3XYZ = 0 ଅଟେ।

ତେଣୁ ଏହି ସମ୍ପୂର୍ଣ୍ଣ ରାଶିର ମାନ ହେବ:

=0= 0

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(vi)

❓ ପ୍ରଶ୍ନ: a6+4a3−1a^6 + 4a^3 - 1

✅ ଉତ୍ତର:

ଏହାକୁ ଆମେ ନିମ୍ନଲିଖିତ ଭାବେ ସଜାଇ ପାରିବା:

=(a2)3+a3+(−1)3−3(a2)(a)(−1)= (a^2)^3 + a^3 + (-1)^3 - 3(a^2)(a)(-1)

ସୂତ୍ର ପ୍ରୟୋଗ କଲେ:

=(a2+a−1)((a2)2+a2+(−1)2−a2(a)−a(−1)−(−1)(a2))= (a^2 + a - 1)((a^2)^2 + a^2 + (-1)^2 - a^2(a) - a(-1) - (-1)(a^2))

=(a2+a−1)(a4+a2+1−a3+a+a2)= (a^2 + a - 1)(a^4 + a^2 + 1 - a^3 + a + a^2)

=(a2+a−1)(a4−a3+2a2+a+1)= (a^2 + a - 1)(a^4 - a^3 + 2a^2 + a + 1)

(vii)

❓ ପ୍ରଶ୍ନ: x3+72−24xx^3 + 72 - 24x

✅ ଉତ୍ତର:

ସଜାଇ ଲେଖିଲେ:

=x3+8+64−24x= x^3 + 8 + 64 - 24x

=x3+23+43−3(x)(2)(4)= x^3 + 2^3 + 4^3 - 3(x)(2)(4)

=(x+2+4)(x2+22+42−2x−8−4x)= (x + 2 + 4)(x^2 + 2^2 + 4^2 - 2x - 8 - 4x)

=(x+6)(x2+4+16−6x−8)= (x + 6)(x^2 + 4 + 16 - 6x - 8)

=(x+6)(x2−6x+12)= (x + 6)(x^2 - 6x + 12)

(viii)

❓ ପ୍ରଶ୍ନ: m6+7m3−8m^6 + 7m^3 - 8

✅ ଉତ୍ତର:

ମଧ୍ୟପଦ ବିଭାଜନ କରିବା:

=m6+8m3−m3−8= m^6 + 8m^3 - m^3 - 8

=m3(m3+8)−1(m3+8)= m^3(m^3 + 8) - 1(m^3 + 8)

=(m3+8)(m3−1)= (m^3 + 8)(m^3 - 1)

=(m3+23)(m3−13)= (m^3 + 2^3)(m^3 - 1^3)

=(m+2)(m2−2m+4)(m−1)(m2+m+1)= (m + 2)(m^2 - 2m + 4)(m - 1)(m^2 + m + 1)

(ix)

❓ ପ୍ରଶ୍ନ: a6+1a6+2(a≠0)a^6 + \frac{1}{a^6} + 2 \quad (a \neq 0)

✅ ଉତ୍ତର:

=(a3)2+(1a3)2+2(a3)(1a3)= (a^3)^2 + \left(\frac{1}{a^3}\right)^2 + 2(a^3)\left(\frac{1}{a^3}\right)

=(a3+1a3)2= \left(a^3 + \frac{1}{a^3}\right)^2

ଏହାକୁ ଆହୁରି ବିଶ୍ଳେଷଣ କଲେ:

=(a+1a)2(a2−1+1a2)2= \left(a + \frac{1}{a}\right)^2 \left(a^2 - 1 + \frac{1}{a^2}\right)^2

(x)

❓ ପ୍ରଶ୍ନ: r6+45r3−8r^6 + 45r^3 - 8

✅ ଉତ୍ତର:

ସଜାଇ ଲେଖିଲେ:

=(r2)3+(3r)3+(−2)3−3(r2)(3r)(−2)= (r^2)^3 + (3r)^3 + (-2)^3 - 3(r^2)(3r)(-2)

=(r2+3r−2)((r2)2+(3r)2+(−2)2−(r2)(3r)−(3r)(−2)−(−2)(r2))= (r^2 + 3r - 2)((r^2)^2 + (3r)^2 + (-2)^2 - (r^2)(3r) - (3r)(-2) - (-2)(r^2))

=(r2+3r−2)(r4+9r2+4−3r3+6r+2r2)= (r^2 + 3r - 2)(r^4 + 9r^2 + 4 - 3r^3 + 6r + 2r^2)

=(r2+3r−2)(r4−3r3+11r2+6r+4)= (r^2 + 3r - 2)(r^4 - 3r^3 + 11r^2 + 6r + 4)

(xi)

❓ ପ୍ରଶ୍ନ: 16x3−54y6−2z3−36xy2z16x^3 - 54y^6 - 2z^3 - 36xy^2z

✅ ଉତ୍ତର:

22 କୁ ସାଧାରଣ ରାଶି ଭାବେ ବାହାରକୁ ଆଣିଲେ:

=2(8x3−27y6−z3−18xy2z)= 2(8x^3 - 27y^6 - z^3 - 18xy^2z)

=2{(2x)3+(−3y2)3+(−z)3−3(2x)(−3y2)(−z)}= 2\{(2x)^3 + (-3y^2)^3 + (-z)^3 - 3(2x)(-3y^2)(-z)\}

=2(2x−3y2−z)((2x)2+(−3y2)2+(−z)2−(2x)(−3y2)−(−3y2)(−z)−(−z)(2x))= 2(2x - 3y^2 - z)((2x)^2 + (-3y^2)^2 + (-z)^2 - (2x)(-3y^2) - (-3y^2)(-z) - (-z)(2x))

=2(2x−3y2−z)(4x2+9y4+z2+6xy2−3y2z+2zx)= 2(2x - 3y^2 - z)(4x^2 + 9y^4 + z^2 + 6xy^2 - 3y^2z + 2zx)

(xii)

❓ ପ୍ରଶ୍ନ: a3+b3−127c3+abca^3 + b^3 - \frac{1}{27}c^3 + abc

✅ ଉତ୍ତର:

=a3+b3+(−c3)3−3(a)(b)(−c3)= a^3 + b^3 + \left(-\frac{c}{3}\right)^3 - 3(a)(b)\left(-\frac{c}{3}\right)

=(a+b−c3)(a2+b2+(−c3)2−ab−b(−c3)−(−c3)a)= \left(a + b - \frac{c}{3}\right)\left(a^2 + b^2 + \left(-\frac{c}{3}\right)^2 - ab - b\left(-\frac{c}{3}\right) - \left(-\frac{c}{3}\right)a\right)

=(a+b−c3)(a2+b2+c29−ab+bc3+ca3)= \left(a + b - \frac{c}{3}\right)\left(a^2 + b^2 + \frac{c^2}{9} - ab + \frac{bc}{3} + \frac{ca}{3}\right)

(xiii)

❓ ପ୍ରଶ୍ନ: 27a3−8b6+125c3+90ab2c27a^3 - 8b^6 + 125c^3 + 90ab^2c

✅ ଉତ୍ତର:

=(3a)3+(−2b2)3+(5c)3−3(3a)(−2b2)(5c)= (3a)^3 + (-2b^2)^3 + (5c)^3 - 3(3a)(-2b^2)(5c)

=(3a−2b2+5c)((3a)2+(−2b2)2+(5c)2−(3a)(−2b2)−(−2b2)(5c)−(5c)(3a))= (3a - 2b^2 + 5c)((3a)^2 + (-2b^2)^2 + (5c)^2 - (3a)(-2b^2) - (-2b^2)(5c) - (5c)(3a))

=(3a−2b2+5c)(9a2+4b4+25c2+6ab2+10b2c−15ca)= (3a - 2b^2 + 5c)(9a^2 + 4b^4 + 25c^2 + 6ab^2 + 10b^2c - 15ca)

(xiv)

❓ ପ୍ରଶ୍ନ: (2x+3)3+(3x−2)3−(5x+1)3(2x + 3)^3 + (3x - 2)^3 - (5x + 1)^3

✅ ଉତ୍ତର:

ଏହାକୁ ଆମେ ଏପରି ଲେଖିପାରିବା:

=(2x+3)3+(3x−2)3+(−5x−1)3= (2x + 3)^3 + (3x - 2)^3 + (-5x - 1)^3

ମନେକର A=2x+3A = 2x + 3, B=3x−2B = 3x - 2, C=−5x−1C = -5x - 1 ।

ଏଠାରେ ଲକ୍ଷ୍ୟ କରନ୍ତୁ: A+B+C=2x+3+3x−2−5x−1=0A + B + C = 2x + 3 + 3x - 2 - 5x - 1 = 0 ।

ଆମେ ଜାଣୁ ଯେ ଯଦି A+B+C=0A + B + C = 0 ହୁଏ, ତେବେ A3+B3+C3=3ABCA^3 + B^3 + C^3 = 3ABC ଅଟେ।

ତେଣୁ ଉତ୍ପାଦକ ହେବ:

=3(2x+3)(3x−2)(−5x−1)= 3(2x + 3)(3x - 2)(-5x - 1)

=−3(2x+3)(3x−2)(5x+1)= -3(2x + 3)(3x - 2)(5x + 1)

📝 6.

❓ ପ୍ରଶ୍ନ: a+b+c=0a + b + c = 0 ହେଲେ ଦର୍ଶାଅ ଯେ, a3+b3+c3=3abca^3 + b^3 + c^3 = 3abc

✅ ସମାଧାନ:

ଆମେ ଜାଣୁ ଯେ ସୂତ୍ର ଅନୁଯାୟୀ:

a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)

ପ୍ରଶ୍ନ ଅନୁସାରେ ଦତ୍ତ ଅଛି ଯେ: a+b+c=0a + b + c = 0 ।

ବର୍ତ୍ତମାନ ଏହି ମୂଲ୍ୟକୁ ସୂତ୍ରରେ ପ୍ରୟୋଗ କଲେ:

a3+b3+c3−3abc=0×(a2+b2+c2−ab−bc−ca)a^3 + b^3 + c^3 - 3abc = 0 \times (a^2 + b^2 + c^2 - ab - bc - ca)

⇒a3+b3+c3−3abc=0\Rightarrow a^3 + b^3 + c^3 - 3abc = 0

⇒a3+b3+c3=3abc(ପ୍ରମାଣିତ)\Rightarrow a^3 + b^3 + c^3 = 3abc \quad \textbf{(ପ୍ରମାଣିତ)}

🧩 7. ❓ ପ୍ରଶ୍ନ: (x−y)3+(y−z)3+(z−x)3(x - y)^3 + (y - z)^3 + (z - x)^3 ର ଉତ୍ପାଦକଗୁଡିକୁ ନିର୍ଣ୍ଣୟ କର ।

✅ ଉତ୍ତର:

ମନେକର a=x−ya = x - y, b=y−zb = y - z, ଏବଂ c=z−xc = z - x ।

ଏଠାରେ ପଦଗୁଡ଼ିକର ଯୋଗଫଳ:

a+b+c=(x−y)+(y−z)+(z−x)=0a + b + c = (x - y) + (y - z) + (z - x) = 0

ଆମେ ଜାଣୁ ଯେ ଯଦି a+b+c=0a + b + c = 0 ହୁଏ, ତେବେ a3+b3+c3=3abca^3 + b^3 + c^3 = 3abc ହେବ।

ତେଣୁ ନିର୍ଣ୍ଣେୟ ଉତ୍ପାଦକ ହେବ:

=3(x−y)(y−z)(z−x)= 3(x - y)(y - z)(z - x)

📝 8. ❓ ପ୍ରଶ୍ନ: ଦର୍ଶାଅ ଯେ, x3+y3+z3−3xyz=12(x+y+z){(x−y)2+(y−z)2+(z−x)2}x^3 + y^3 + z^3 - 3xyz = \frac{1}{2}(x + y + z)\{(x - y)^2 + (y - z)^2 + (z - x)^2\}

✅ ସମାଧାନ:

ବାମପକ୍ଷ:

=x3+y3+z3−3xyz= x^3 + y^3 + z^3 - 3xyz

ସୂତ୍ର ଅନୁଯାୟୀ ଏହାକୁ ସମ୍ପ୍ରସାରଣ କଲେ:

=(x+y+z)(x2+y2+z2−xy−yz−zx)= (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)

ବର୍ତ୍ତମାନ ଉଭୟ ଉପରେ ଏବଂ ତଳେ 22 ଦ୍ୱାରା ଗୁଣନ ଓ ଭାଗ କରିବା:

=12(x+y+z)(2x2+2y2+2z2−2xy−2yz−2zx)= \frac{1}{2}(x + y + z)(2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx)

ବର୍ତ୍ତମାନ କୁଟିଳ ବନ୍ଧନୀ ମଧ୍ୟରେ ଥିବା ପଦଗୁଡ଼ିକୁ ସଜାଇ ଲେଖିବା:

=12(x+y+z){(x2−2xy+y2)+(y2−2yz+z2)+(z2−2zx+x2)}= \frac{1}{2}(x + y + z)\{(x^2 - 2xy + y^2) + (y^2 - 2yz + z^2) + (z^2 - 2zx + x^2)\}

ବର୍ଗ ସୂତ୍ର (a−b)2(a - b)^2 ପ୍ରୟୋଗ କଲେ:

=12(x+y+z){(x−y)2+(y−z)2+(z−x)2}= \frac{1}{2}(x + y + z)\{(x - y)^2 + (y - z)^2 + (z - x)^2\}

= ଦକ୍ଷିଣପକ୍ଷ (ପ୍ରମାଣିତ)