The coefficient of x^{-5} in the binomial exp | vedclass

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Question

(A) Simplify the expression inside the bracket: Let $u = x^{1/3}$ and $v = x^{1/2}$. The first term is $\frac{u^3 + 1}{u^2 - u + 1} = \frac{(u+1)(u^2

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$1$

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(A) Simplify the expression inside the bracket: Let $u = x^{1/3}$ and $v = x^{1/2}$. The first term is $\frac{u^3 + 1}{u^2 - u + 1} = \frac{(u+1)(u^2 - u + 1)}{u^2 - u + 1} = u + 1 = x^{1/3} + 1$. The second term is $\frac{v^2 - 1}{v(v - 1)} = \frac{(v-1)(v+1)}{v(v-1)} = \frac{v+1}{v} = 1 + \frac{1}{v} = 1 + x^{-1/2}$. Subtracting the two terms: $(x^{1/3} + 1) - (1 + x^{-1/2}) = x^{1/3} - x^{-1/2}$. Now,we need the coefficient of $x^{-5}$ in $(x^{1/3} - x^{-1/2})^{10}$. The general term $T_{r+1} = {^{10}C_r} (x^{1/3})^{10-r} (-x^{-1/2})^r = {^{10}C_r} (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$. Set the exponent equal to $-5$: $\frac{10-r}{3} - \frac{r}{2} = -5$. Multiply by $6$: $2(10-r) - 3r = -30 \implies 20 - 2r - 3r = -30 \implies 50 = 5r \implies r = 10$. The coefficient is ${^{10}C_{10}} (-1)^{10} = 1 	imes 1 = 1$.

The coefficient of $x^{-5}$ in the binomial expansion of ${\left( {\frac{{x + 1}}{{{x^{2/3}} - {x^{1/3}} + 1}} - \frac{{x - 1}}{{x - {x^{1/2}}}}} \right)^{10}}$ where $x \ne 0, 1$,is

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