Question

1. $\frac{x^{3}-64}{x^{3}+64}div\frac{x^{2}-16}{x^{2}-4x + 16}$

Explanation:

Step1: Use difference - of - cubes and sum - of - cubes formulas

We know that $a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})$ and $a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})$. So, $x^{3}-64=(x - 4)(x^{2}+4x + 16)$ and $x^{3}+64=(x + 4)(x^{2}-4x + 16)$. Also, $x^{2}-16=(x + 4)(x - 4)$.

Step2: Rewrite the division as multiplication

Dividing by a fraction is the same as multiplying by its reciprocal. So, $\frac{x^{3}-64}{x^{3}+64}\div\frac{x^{2}-16}{x^{2}-4x + 16}=\frac{x^{3}-64}{x^{3}+64}\times\frac{x^{2}-4x + 16}{x^{2}-16}$.

Step3: Substitute the factorized forms

Substitute the factorized expressions: $\frac{(x - 4)(x^{2}+4x + 16)}{(x + 4)(x^{2}-4x + 16)}\times\frac{x^{2}-4x + 16}{(x + 4)(x - 4)}$.

Step4: Cancel out common factors

Cancel out the common factors $(x - 4)$ and $(x^{2}-4x + 16)$: $\frac{x^{2}+4x + 16}{(x + 4)^{2}}$.

Answer:

$\frac{x^{2}+4x + 16}{(x + 4)^{2}}$