Question

How do I solve the cube root of -(x-7)^4? PLS EXPLAIN

Answer 1

Renmember that
$\sqrt[3]{ab}=( \sqrt[3]{a})( \sqrt[3]{b})$
and
$\sqrt[n]{x^m} =x^ \frac{m}{n}$

split it up

$\sqrt[3]{-1(x-7)^4} =( \sqrt[3]{-1} )( \sqrt[3]{(x-7)^4})$=
$(-1)((x-7)^ \frac{4}{3})=(-1)(x-7)(x-7)^ \frac{1}{3}=-(x-7) \sqrt[3]{x-7}$=
$(7-x) \sqrt[3]{x-7}=7\sqrt[3]{x-7}-x\sqrt[3]{x-7}$