Question

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

Answer 1

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

Formula used: (i)

ⁿC_r = \(\frac{n!}{(n-r)!(r)!}\)

(ii) (a+b)ⁿ = ⁿC₀aⁿ + ⁿC₁a^n-1b + ⁿC₂a^n-2b² + … +ⁿC_n-1ab^n-1 + ⁿC_nbⁿ 

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