Question

x³-64 in prime factored form​

Answer 1

$\bf \textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) \qquad \qquad a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ x^3-64\implies x^3-4^3\implies (x-4)(x^2+4x+4^2)\implies (x-4)(x^2+4x+16)$