## College Algebra (6th Edition)

$(x+4)(x^{2}-4x+16)$
If a and b are real numbers, we know that $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2)}$. Therefore, we know that $x^{3}+64=(x+4)(x^{2}-x\times4+4^{2})=(x+4)(x^{2}-4x+16)$. In this case, $a=x$ and $b=4$.