Q. 6

# Let’s resolve int

(i) 1000a3+27b6

= (10a)3 + (3b2)6

Using x3 + y3 = (x = y)(x2 - xy + y2)

= (10a + 3b2)[(10a)2 - (10a)(3b2) + (3b2)2]

= (10a + 3b2)(100a2 - 30ab2 + 9b4)

(ii) 1-216 z3

= 1 – (6z)3

Using x3 – y3 = (x – y)(x2 + xy + y2)

= (1 – 6z)(12 + 1(6z) + (6z)2)

= (1 – 6z)(1 + 6z + 36z2)

(iii) m4 - m

= m(m3 – 1)

= m(m3 – 13)

Using x3 – y3 = (x – y)(x2 + xy + y2)

= m(m – 1)(m2 + m + 1)

(iv) 192a3+3

= 3(64a3 + 1)

= 3[(4a)3 + 13]

Using x3 + y3 = (x + y)(x2 – xy + y2)

= 3(4a + 1)[(4a)2 – 4a + 1]

= 3(4a + 1)(16a2 – 4a + 1)

(v) 16a4x3 + 54ay3

= 2a(8a3x3 + 27y3)

= 2a[(2ax)3 + (3y)3]

Using x3 + y3 = (x + y)(x2 – xy + y2)

= 2a(2ax + 3y)[(2ax)2 – 2ax(3y) + (3y)2]

= 2a(2ax + 3y)(4a2x2 – 6axy + 9y2)

(vi) 729a3 b3 c3-125

= (9abc)3 - 53

= (9abc – 5)[(9abc)2 + 9abc(5) + 52]

= (9abc – 5)(81a2b2c2 + 45abc + 25)

(vii)

Using x3 - y3 = (x - y)(x2 + xy + y2)

(viii)

Using x3 – y3 = (x – y)(x2 + xy + y2)

(ix)x3 + 3x2y + 3xy2 + 2y3

= x3 + y3 + 3x2y + 3xy2 + y3

= (x + y)3 + y3

[ a3 + b3 + 3a2b + 3ab2 = (a + b)3]

= (x + y + y)[(x + y)2 - (x + y)y + y2]

[Using x3 – y3 = (x + y)(x2 - xy + y2)]

= (x + 2y)(x2 + y2 - xy - y2 + y2)

= (x + 2y)(x2 – xy + y2)

(x) 1 + 9x + 27x2 + 28x3

= 27x3 + 1 + 27x2 + 9x + x3

= (3x)3 + 13 + 3(3x)3(1) + 3(3x)(1)2

= (3x + 1)3 + x3

[ a3 + b3 + 3a2b + 3ab2 = (a + b)3]

Now, Using x3 + y3 = (x + y)(x2 - xy + y2)

= (3x + 1 + x)[(3x + 1)2 - (3x + 1)x + x2]

= (4x + 1)(9x2 + 6x + 1 - 3x2 - x + x2)

= (4x + 1)(7x2 + 5x + 1)

(xi) x3 – 9y3 – 3xy(x-y)

x3 – y3 – 3xy (x – y) – 8y3

[ x3 – y3 = x3 – y3 – 3xy (x – y)]

= (x – y)3 – (2y)3

= (x – y – (2y)) [(x – y)2 + (x – y) (2y) + (2y)2]

[Using x3 + y3 = (x + y) (x2 – xy + y2)]

= (x -3 y) (x2 + y2 – 2xy + 2xy – 2y2 + 4y2)

= (x - 3y) (x2 + 3y2)

(xii)8 – a3 + 3a2b – 3ab2 + b3

= b3 – a3 – 3ab2 + 3a2b + 8

Now, As (x + y)3 = x3 + y3 + 3x2y + 3xy2

= (b – a)3 + 23

Using x3 + y3 = (x + y) (x2 - xy + y2)

= (b – a + 2) [(b – a)2 - (b – a)2 + 22]

= (b – a + 2) (b2 + a2 – 2ab - 2b + 2a + 4)

(xiii) x6+3x4 b2+3x2 b4+b6+a3b3

= (x2)3 + (b2)3 + 3(x2)2(b2) + 3(x2)(b2)2 + a3b3

As (x + y)3 = x3 + y3 + 3x2y + 3xy2

= (x2 + b2)3 + (ab)3

Using x3 + y3 = (x + y)(x2 – xy + y2)

= (x2 + b2 + ab)[(x2 + b2)2 - (x2 + b2)ab + a2b2]

(xiv) x6 + 27

= (x2)3 + (3)3

Using x3 + y3 = (x + y)(x2 – xy + y2)

= (x2 + 3)[(x2)2 - 3x2 + 32)

= (x2 + 3)(x4 - 3x2 + 9)

(xv)x6 – y6

= (x3)2 – (y3)2

Using (a2 – b2) = (a – b)(a + b)

= (x3 – y3)(x3 + y3)

Using x3 – y3 = (x – y)(x2 + xy + y2) and

x3 + y3 = (x + y)(x2 – xy + y2)

= (x – y)(x2 + xy + y2)(x + y)(x2 – xy + y2)

(xvi) x12 – y12

= (x6)2 – (y6)2

Using (a2 – b2) = (a – b)(a + b)

= (x6 – y6)(x6 + y6)

= [(x3)2 – (y3)2 �][(x2)3 + (y2)3]

Using (a2 – b2) = (a – b)(a + b)

= (x3 – y3)(x3 + y3)[(x2)3 + (y2)3]

Using x3 – y3 = (x – y)(x2 + xy + y2) and

x3 + y3 = (x + y)(x2 – xy + y2)

= (x – y)(x2 + xy + y2)(x + y)(x2 – xy + y2)(x2 + y2)(x4 - x2y2 + x6y6)

(xvii) m3 -n3-m(m2-n2 )+n(m-n)2

Using x3 – y3 = (x – y)(x2 + xy + y2) and

Using (a2 – b2) = (a – b)(a + b)

= (m – n)(m2 + mn + n2) – m(m – n)(m + n) + n(m – n)2

Taking (m – n) as common, we get

= (m – n)[m2 + mn + n2 – m(m + n) + n(m – n)]

= (m – n)(m2 + mn + n2 – m2 – mn + mn – n2)

= (m – n)mn

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