Using binomial theorem, expand each of the following:

Solution:

To find: Expansion of $\left(1+2 x-3 x^{2}\right)^{4}$

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, $\left(1+2 x-3 x^{2}\right)^{4}$

Let $(1+2 x)=a$ and $\left(-3 x^{2}\right)=b \ldots$ (i)

Now the equation becomes $(a+b)^{4}$

$\Rightarrow\left[{ }^{4} \mathrm{C}_{0}(\mathrm{a})^{4-0}\right]+\left[{ }^{4} \mathrm{C}_{1}(\mathrm{a})^{4-1}(\mathrm{~b})^{1}\right]+\left[{ }^{4} \mathrm{C}_{2}(\mathrm{a})^{4-2}(\mathrm{~b})^{2}\right]+\left[{ }^{4} \mathrm{C}_{3}(\mathrm{a})^{4-3}(\mathrm{~b})^{3}\right]+\left[{ }^{4} \mathrm{C}_{4}(\mathrm{~b})^{4}\right]$

$\Rightarrow\left[{ }^{4} \mathrm{C}_{0}(\mathrm{a})^{4}\right]+\left[{ }^{4} \mathrm{C}_{1}(\mathrm{a})^{3}(\mathrm{~b})^{1}\right]+\left[{ }^{4} \mathrm{C}_{2}(\mathrm{a})^{2}(\mathrm{~b})^{2}\right]+\left[{ }^{4} \mathrm{C}_{3}(\mathrm{a})(\mathrm{b})^{3}\right]+\left[{ }^{4} \mathrm{C}_{4}(\mathrm{~b})^{4}\right]$

(Substituting value of b from eqn. i)

$\Rightarrow\left[\frac{4 !}{0 !(4-0) !}(a)^{4}\right]+\left[\frac{4 !}{1 !(4-1) !}(a)^{3}\left(-3 x^{2}\right)^{1}\right]+\left[\frac{4 !}{2 !(4-2) !}(a)^{2}\left(-3 x^{2}\right)^{2}\right]$

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

(Substituting value of b from eqn. i)

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

$\left[4(1+2 x)\left(27 x^{6}\right)^{3}\right]+\left[1\left(81 x^{8}\right)^{4}\right]$

We need the value of $a^{4}, a^{3}$ and $a^{2}$, where $a=(1+2 x)$ ........(ii)

For $(1+2 x)^{4}$, Applying Binomial theorem

$(1+2 x)^{4} \Rightarrow$

${ }^{4} C_{0}(1)^{4-0}+{ }^{4} C_{1}(1)^{4-1}(2 x)^{1}+{ }^{4} C_{2}(1)^{4-2}(2 x)^{2}+{ }^{4} C_{3}(1)^{4-3}(2 x)^{3}+{ }^{4} C_{4}(2 x)^{4}$

$\Rightarrow \frac{4 !}{0 !(4-0) !}(1)^{4}+\frac{4 !}{1 !(4-1) !}(1)^{3}(2 x)^{1}+\frac{4 !}{2 !(4-2) !}(1)^{2}(2 x)^{2}$

$+\frac{4 !}{3 !(4-3) !}(1)(2 x)^{3}+\frac{4 !}{4 !(4-4) !}(2 x)^{4}$

$\Rightarrow[1]+[4(1)(2 x)]+\left[6(1)\left(4 x^{2}\right)\right]+\left[4(1)\left(8 x^{3}\right)\right]+\left[1\left(16 x^{4}\right)\right]$

$\Rightarrow 1+8 x+24 x^{2}+32 x^{3}+16 x^{4}$

We have $(1+2 x)^{4}=1+8 x+24 x^{2}+32 x^{3}+16 x^{4} \ldots$ (iii)

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

For, $(1+2 x)^{3}$, substituting $a=1$ and $b=2 x$ in the above formula

$\Rightarrow 1^{3}+(2 x)^{3}+3(1)^{2}(2 x)+3(1)(2 x)^{2}$

$\Rightarrow 1+8 x^{3}+6 x+12 x^{2}$

$\Rightarrow 8 x^{3}+12 x^{2}+6 x+1 \ldots$ (iv)

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

For, $(1+2 x)^{2}$, substituting $a=1$ and $b=2 x$ in the above formula

$\Rightarrow(1)^{2}+2(1)(2 x)+(2 x)^{2}$

$\Rightarrow 1+4 x+4 x^{2}$

$\Rightarrow 4 x^{2}+4 x+1 \ldots(v)$

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

$\Rightarrow 1\left(1+8 x+24 x^{2}+32 x^{3}+16 x^{4}\right)-4\left(8 x^{3}+12 x^{2}+6 x+1\right)\left(3 x^{2}\right)$

$+6\left(4 x^{2}+4 x+1\right)\left(9 x^{4}\right)-4(1+2 x)\left(27 x^{6}\right)^{3}+1\left(81 x^{8}\right)$

$\Rightarrow 1\left(1+8 x+24 x^{2}+32 x^{3}+16 x^{4}\right)-4\left(24 x^{5}+36 x^{4}+18 x^{3}+3 x^{2}\right)$

$+6\left(36 x^{6}+36 x^{5}+9 x^{4}\right)-4\left(27 x^{6}+54 x^{7}\right)+1\left(81 x^{8}\right)$

$\Rightarrow 1+8 x+24 x^{2}+32 x^{3}+16 x^{4}-96 x^{5}-144 x^{4}-72 x^{3}-12 x^{2}+216 x^{6}+216 x^{5}+54 x^{4}-$

$108 x^{6}-216 x^{7}+81 x^{8}$

On rearranging

Ans) $81 x^{8}-216 x^{7}+108 x^{6}+120 x^{5}-74 x^{4}-40 x^{3}+12 x^{2}+8 x+1$