Binomial Theorem

To determine the absolute value of the difference between the coefficients of $x^4$ and $x^6$ in the expansion of the given expression, we start with:$x^{2}-2x^{2}+(x+1)^{4}(x^{2}-1)^{2}$We simplify the function as follows:$f(x)=\frac{2x^{2}}{(x^{2}+1)(x^{2}+2)}$Expanding, we have:$f(x)=2x^{2}(x^{2}+1)^{-1}(x^{2}+2)^{-1}$Simplifying further gives:$f(x)=\frac{2x^{2}}{2}(1+x^{2})^{-1}\left(1+\frac{x^{2}}{2}\right)^{-1}$This can be expanded into:$f(x)=x^{2}\left[1-x^{2}+x^{4}-x^{6}+\cdots\right]\cdot\left[1-\frac{x^{2}}{2}+\frac{x^{4}}{4}-\frac{x^{4}}{8}+\cdots\right]$Now, let's find the coefficients.For $x^4$ :Coefficient calculation:$\left[-\frac{1}{2}-1\right]=\frac{-3}{2}$For $x^6$ :Coefficient calculation:$\left[\frac{1}{4}+\frac{1}{2}+1\right]-\frac{7}{4}$Now calculate the absolute difference:$\text{Difference}=\left|\frac{7}{4}-\left(\frac{-3}{2}\right)\right|=\left|\frac{7}{4}+\frac{3}{2}\right|$$=\left|\frac{7}{4}+\frac{6}{4}\right|=\left|\frac{13}{4}\right|=\frac{13}{4}$This process allows us to correctly determine the absolute value of the difference between the coefficients of $x^4$ and $x^6$ in the expansion.