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

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

Formula used: (i)

(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

Substituting the value of p from eqn. (i)

… (ii)

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

27x6 – 8a3x3 – 54ax5 + 36a2x4 … (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

9x4 – 12x3a + 4a2x2 … (iv)

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

27x6 – 8a3x3 – 54ax5 + 36a2x4 + 81a2x4 – 108x3a3 + 36a4x2 + 81a4x2 – 54a5x + 27a6

On rearranging

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

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