Answer
$\dfrac{1-x^{2}}{x^{3}-1}=-\dfrac{x+1}{x^{2}+x+1}$
Work Step by Step
$\dfrac{1-x^{2}}{x^{3}-1}$
We factor the numerator using the difference of squares factoring formula, which is: $a^{2}-b^{2}=(a-b)(a+b)$.
Also, we factor the denominator using the difference of cubes factoring formula, which is: $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$
$\dfrac{1-x^{2}}{x^{3}-1}=\dfrac{(1-x)(1+x)}{(x-1)(x^{2}+x+1)}=...$
We are allowed to change the sign of the numerator or denominator if we also change the sign of the fraction. Knowing that, we rewrite the expression and simplify:
$...=-\dfrac{-(1-x)(1+x)}{(x-1)(x^{2}+x+1)}=-\dfrac{(x-1)(1+x)}{(x-1)(x^{2}+x+1)}=...$
$...=-\dfrac{x+1}{x^{2}+x+1}$
