Question

Simplify: $2\sqrt[3]{316}+\sqrt[3]{381}-\sqrt[3]{3128}+\sqrt[3]{3192}$.

Answer 1

Reduce the surds to their simplest form, we get,
$2\sqrt[3]{316}=2\sqrt[3]{38\times 2}=2\times 2\sqrt[3]{32}=4\sqrt[3]{32},\sqrt[3]{381}=\sqrt[3]{327\times 3}=3\sqrt[3]{33},$
$\sqrt[3]{3128}=\sqrt[3]{364\times 2}=\sqrt[3]{32},\sqrt[3]{3192}=\sqrt[3]{364\times 3}=4\sqrt[3]{33}.$

$\therefore 2\sqrt[3]{316}+\sqrt[3]{381}-\sqrt[3]{3128}+\sqrt[3]{3192}=4\sqrt[3]{32}+3\sqrt[3]{32}-4\sqrt[3]{32}+4\sqrt[3]{33}$
$=4\sqrt[3]{32}-4\sqrt[3]{32}+3\sqrt[3]{33}+4\sqrt[3]{33}$ (by re-arranging the like terms)
$=(3+4)\sqrt[3]{33}=7\sqrt[3]{33}$ .

$\therefore 2\sqrt[3]{316}+\sqrt[3]{381}-\sqrt[3]{3128}+\sqrt[3]{3192}=7\sqrt[3]{33}$ .