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

The task is to locate the coefficient of the term whose power of $$x$$ is $$-5$$ in the expansion of

$$\left(\dfrac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\dfrac{x-1}{x-x^{\frac12}}\right)^{10},\qquad x\neq0,\;x\neq1.$$

First the two fractions inside will be simplified separately.

For the first fraction let $$y=x^{\frac13}.$$ Then

$$x^{\frac23}=y^{2},\qquad x+1=y^{3}+1.$$

The fraction becomes

$$\dfrac{y^{3}+1}{\,y^{2}-y+1\,}.$$

Using the factorisation $$y^{3}+1=(y+1)(y^{2}-y+1)$$ gives

$$\dfrac{(y+1)(y^{2}-y+1)}{\,y^{2}-y+1\,}=y+1.$$

Undoing the substitution, the first fraction is

$$x^{\frac13}+1.$$

For the second fraction put $$z=x^{\frac12}.$$ Then $$x=z^{2},$$ so

$$x-1=z^{2}-1=(z-1)(z+1),$$ $$x-x^{\frac12}=z^{2}-z=z(z-1).$$

Hence

$$\dfrac{z^{2}-1}{\,z^{2}-z\,}= \dfrac{(z-1)(z+1)}{\,z(z-1)\,}= \dfrac{z+1}{z}=1+\dfrac1z.$$

Returning to $$x,$$ the second fraction is

$$1+x^{-\frac12}.$$

The expression inside the large parentheses therefore simplifies to

$$(x^{\frac13}+1)-(1+x^{-\frac12})=x^{\frac13}-x^{-\frac12}.$$

The whole expression becomes

$$(x^{\frac13}-x^{-\frac12})^{10}.$$

Denote $$a=x^{\frac13},\;b=-x^{-\frac12};$$ then the binomial theorem gives the general term

$$T_{k}= \binom{10}{k}a^{\,10-k}b^{\,k}= \binom{10}{k}(x^{\frac13})^{10-k}\bigl(-x^{-\frac12}\bigr)^{k},\qquad k=0,1,\dots,10.$$

The power of $$x$$ in this term is

$$\left(10-k\right)\left(\dfrac13\right)+k\!\left(-\dfrac12\right)=\dfrac{10-k}{3}-\dfrac{k}{2}.$$

Putting everything over the common denominator 6,

$$\dfrac{2(10-k)-3k}{6}= \dfrac{20-2k-3k}{6}= \dfrac{20-5k}{6}.$$

We want this exponent to equal $$-5,$$ so set

$$\dfrac{20-5k}{6}=-5.$$

Multiplying by 6:

$$20-5k=-30.$$

Subtract 20:

$$-5k=-50.$$

Divide by $$-5$$:

$$k=10.$$

Only the term with $$k=10$$ contributes to the coefficient of $$x^{-5}.$$ For this value,

$$T_{10}= \binom{10}{10}(x^{\frac13})^{0}\bigl(-x^{-\frac12}\bigr)^{10} =1\cdot1\cdot\bigl[(-1)^{10}x^{-5}\bigr]=x^{-5}.$$

The numerical coefficient is therefore $$1.$$

Hence, the correct answer is Option C.