Question

Show that $\sqrt[3]{3216\times (-343)}=\sqrt[3]{3216}\times \sqrt[3]{3-343}$

Answer 1

$\sqrt[3]{3216\times (-343)}=\sqrt[3]{3216}\times \sqrt[3]{3-343}$
$\sqrt[3]{3{\left(6\right)}^3\times {(-7)}^3}=\sqrt[3]{3{\left(6\right)}^3}\times \sqrt[3]{3{(-7)}^3}$
$6\times -7=6\times -7$
$-42=-42$
$-42+42=0$