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Using binomial theorem, expand each of the following: (3x2 – 2ax + 3a2)3

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To find: Expansion of (3x2 – 2ax + 3a2)3

Formula used: (i)

nCr\(\frac{n!}{(n-r)!(r)!}\)

(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + … +nCn-1abn-1 + nCnbn 

We have, (3x2 – 2ax + 3a2)3 

Let, (3x2 – 2ax) = p … (i) 

The equation becomes (p + 3a2)3

We need the value of p3 and p2 , where p = 3x2 – 2ax 

For, (a+b)3 , we have formula a3+b3+3a2b+3ab2 

For, (3x2 – 2ax)3 , substituting a = 3x2 and b = –2ax in the above formula 

⇒[(3x2)3] + [(-2ax)3] + [3(3x2)2(-2ax)] + [3(3x2)(-2ax)2]

⇒ 27x6 – 8a3x3 – 54ax5 + 36a2x 4 … (iii) 

For, (a+b)2 , we have formula a2+2ab+b2 

For, (3x2 – 2ax)3 , substituting a = 3x2 and b = –2ax in the above formula 

⇒{(3x2)2] + [2(3x2)(-2ax)] +[(-2ax)2]

⇒ 9x4 – 12x3a + 4a2x2 … (iv) 

Putting the value obtained from eqn. (iii) and (iv) in eqn. (ii)

⇒ 27x6 – 8a3x3 – 54ax5 + 36a2x4 + 81a2x4 – 108x3a 3 + 36a4x2 + 81a4x2 – 54a5x + 27a6 

On rearranging

27x6 – 54ax5 + 117a2x4 – 116x3a3 + 117a4x2 – 54a5x + 27a6

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