what is the completely factored form of $x^{3}-64x$?\n$x(x - 8)(x - 8)$\n$(x - 4)(x^{2}+4x + 16)$\n$x(x…

Explanation:

Step1: Factor out the common factor

First, factor out the common factor $x$ from $x^{3}-64x$. We get $x(x^{2}-64)$.

Step2: Use the difference - of - squares formula

The expression $x^{2}-64$ is a difference of squares since $x^{2}-64=x^{2}-8^{2}$. According to the difference - of - squares formula $a^{2}-b^{2}=(a - b)(a + b)$, where $a = x$ and $b = 8$, so $x^{2}-64=(x - 8)(x + 8)$.

Step3: Write the completely factored form

The completely factored form of $x^{3}-64x$ is $x(x - 8)(x + 8)$.

Answer:

C. $x(x - 8)(x + 8)$