Question

Simplify the binomial ${(\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}})}^{10}$ and find the term of its expansion which does not contain $x$.

Answer 1

$(\frac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{x-1}{x-x^{\frac{1}{2}}}{)}^{10}$

$=[\frac{(x+1)(x^{\frac{1}{3}}+1)}{(x^{\frac{1}{3}}+1)(x^{\frac{2}{3}}-x^{\frac{1}{3}}+1)}$ $-\frac{(x^{\frac{1}{2}}{)}^2-1}{x^{\frac{1}{2}}(x^{\frac{1}{2}}-1)}{]}^{10}$

$=[\frac{(x+1)(x^{\frac{1}{3}}+1)}{(x^{\frac{1}{3}}{)}^3+1}-\frac{(x^{\frac{1}{2}}-1)(x^{\frac{1}{2}}+1)}{x^{\frac{1}{2}}(x^{\frac{1}{2}}-1)}{]}^{10}$

$=[\frac{(x+1)(x^{\frac{1}{3}}+1)}{(x+1)}-\frac{x^{\frac{1}{2}}+1}{x^{\frac{1}{2}}}{]}^{10}$

$=(x^{\frac{1}{3}}+1-1-\frac{1}{\sqrt{x}}{)}^{10}$

$=(\sqrt[3]{3x}-\frac{1}{\sqrt{x}}{)}^{10}$

Let $a^{th}$ term in the expansion does not contain $x$,

$a^{th}term{=}^n{C}_{a-1}{x}^{\frac{a}{3}}\frac{1}{(x{)}^{\frac{10-a}{2}}}$

So, $\frac{a}{3}-\frac{10-a}{2}=0$

$\Rightarrow 2a-30+3a=0$

$\Rightarrow 5a=30$

$\Rightarrow a=6$