Mathematics - Binomial Theorem Question with Solution | TestHub

Given the expression:

$$=\frac{{\left(x^{1/3}\right)}^3+(1{)}^3}{x^{2/3}-x^{1/3}+1}-\frac{(x-1)}{x^{1/2}\left(x^{1/2}-1\right)}$$

This simplifies to:

$$=\frac{\left(x^{1/3}+1\right)\left(x^{2/3}-x^{1/3}+1\right)}{\left(x^{2/3}-x^{1/3}+1\right)}-\frac{\left(x^{1/2}+1\right)\left(x^{1/2}-1\right)}{x^{1/2}\left(x^{1/2}-1\right)}$$

$$=\left(x^{1/3}+1\right)-\left(1+x^{-1/2}\right)=x^{1/3}-x^{-1/2}$$

Therefore, we need to evaluate ${\left(\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}\right)}^{10}$.

Substituting the simplified expression, we get:

$$={\left(x^{1/3}-x^{-1/2}\right)}^{10}$$

To find the term independent of $x$, we use the general term $T_{r+1}$ in the expansion of $(a+b{)}^n$, which is given by $T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$.

For ${\left(x^{1/3}-x^{-1/2}\right)}^{10}$, the $(r+1{)}^{th}$ term is:

$$T_{r+1}={}^{10}{C}_r{\left(x^{1/3}\right)}^{10-r}\cdot (-1{)}^r\cdot {\left(x^{-1/2}\right)}^r$$

$$=(-1{)}^r{}^{10}{C}_r{x}^{\left(\frac{10-r}{3}-\frac{r}{2}\right)}$$

For the term to be independent of $x$, the exponent of $x$ must be zero:

$$\frac{10-r}{3}-\frac{r}{2}=0$$

Multiplying by 6 to clear the denominators:

$$2(10-r)-3r=0$$

$$20-2r-3r=0$$

$$20-5r=0$$

$$5r=20$$

$$r=4$$

Hence, the required coefficient is ${}^{10}{C}_4(-1{)}^4$.

$${}^{10}{C}_4=\frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}=10\times 3\times 7=210$$

So, the coefficient is $210\times 1=210$.