Using binomial theorem, expand each of the following:

To find: Expansion of $\left(3 x^{2}-2 a x+3 a^{2}\right)^{3}$

Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$

(ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$

We have, $\left(3 x^{2}-2 a x+3 a^{2}\right)^{3}$

Let, $\left(3 x^{2}-2 a x\right)=p \ldots$ (i)

The equation becomes $\left(p+3 a^{2}\right)^{3}$

$\Rightarrow\left[{ }^{3} C_{0}(p)^{3-0}\right]+\left[{ }^{3} C_{1}(p)^{3-1}\left(3 a^{2}\right)^{1}\right]+\left[{ }^{3} C_{2}(p)^{3-2}\left(3 a^{2}\right)^{2}\right]+\left[{ }^{3} C_{3}\left(3 a^{2}\right)^{3}\right]$

$\Rightarrow\left[{ }^{3} C_{0}(p)^{3}\right]+\left[{ }^{3} C_{1}(p)^{2}\left(3 a^{2}\right)\right]+\left[{ }^{3} C_{2}(p)\left(9 a^{4}\right)\right]+\left[{ }^{3} C_{3}\left(27 a^{6}\right)\right]$

Substituting the value of p from eqn. (i)

$\Rightarrow\left[\frac{3 !}{0 !(3-0) !}\left(3 x^{2}-2 a x\right)^{3}\right]+\left[\frac{3 !}{1 !(3-1) !}\left(3 x^{2}-2 a x\right)^{2}\left(3 a^{2}\right)\right]$

$+\left[\frac{3 !}{2 !(3-2) !}\left(3 x^{2}-2 a x\right)\left(9 a^{4}\right)\right]+\left[\frac{3 !}{3 !(3-3) !}\left(27 a^{6}\right)\right]$

$\Rightarrow\left[1\left(3 x^{2}-2 a x\right)^{3}\right]+\left[3\left(3 x^{2}-2 a x\right)^{2}\left(3 a^{2}\right)\right]+\left[3\left(3 x^{2}-2 a x\right)\left(9 a^{4}\right)\right]+$

$\left[1\left(27 a^{6}\right)^{3}\right]$

(ii)

We need the value of $p^{3}$ and $p^{2}$, where $p=3 x^{2}-2 a x$

For, $(a+b)^{3}$, we have formula $a^{3}+b^{3}+3 a^{2} b+3 a b^{2}$

For, $\left(3 x^{2}-2 a x\right)^{3}$, substituting $a=3 x^{2}$ and $b=-2 a x$ in the above formula

$\Rightarrow\left[\left(3 x^{2}\right)^{3}\right]+\left[(-2 a x)^{3}\right]+\left[3\left(3 x^{2}\right)^{2}(-2 a x)\right]+\left[3\left(3 x^{2}\right)(-2 a x)^{2}\right]$

$\Rightarrow 27 x^{6}-8 a^{3} x^{3}-54 a x^{5}+36 a^{2} x^{4} \ldots$ (iii)

For, $(a+b)^{2}$, we have formula $a^{2}+2 a b+b^{2}$

For, $\left(3 x^{2}-2 a x\right)^{3}$, substituting $a=3 x^{2}$ and $b=-2 a x$ in the above formula

$\Rightarrow\left[\left(3 x^{2}\right)^{2}\right]+\left[2\left(3 x^{2}\right)(-2 a x)\right]+\left[(-2 a x)^{2}\right]$

$\Rightarrow 9 x^{4}-12 x^{3} a+4 a^{2} x^{2} \ldots$ (iv)

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

$\Rightarrow\left[1\left(27 x^{6}-8 a^{3} x^{3}-54 a x^{5}+36 a^{2} x^{4}\right)\right]+$

$\left[3\left(9 x^{4}-12 x^{3} a+4 a^{2} x^{2}\right)\left(3 a^{2}\right)\right]+\left[3\left(3 x^{2}-2 a x\right)\left(9 a^{4}\right)\right]+\left[1\left(27 a^{6}\right)\right]$

$\Rightarrow 27 x^{6}-8 a^{3} x^{3}-54 a x^{5}+36 a^{2} x^{4}+81 a^{2} x^{4}-108 x^{3} a^{3}+36 a^{4} x^{2}+81 a^{4} x^{2}-54 a^{5} x+27 a^{6}$

On rearranging

Ans) $27 x^{6}-54 a x^{5}+117 a^{2} x^{4}-116 x^{3} a^{3}+117 a^{4} x^{2}-54 a^{5} x+27 a^{6}$