melodykap

2021-07-30

The simplest form of the expression, $\sqrt[3]{40{x}^{4}{y}^{7}}$

BleabyinfibiaG

Given: $\sqrt[3]{40{x}^{4}{y}^{7}}$
Explanation:
We will make factors that have powers of 3 and its multiples because the index is 3.
Identifying perfect cube factors, we will get:
$\sqrt[3]{40{x}^{4}{y}^{7}}=\sqrt[3]{5×8×{x}^{3}×x×{y}^{6}×y}$
Grouping the perfect cubes, we get:
$\sqrt[3]{40{x}^{4}{y}^{7}}=\sqrt[3]{\left(8{x}^{3}{y}^{6}\right)\left(5xy\right)}$
Factoring into two radicals, we get:
$\sqrt[3]{40{x}^{4}{y}^{7}}=\sqrt[3]{\left(8{x}^{3}{y}^{6}\right)}×\sqrt[3]{\left(5xy\right)}$
$\sqrt[3]{8}=2$ and $\sqrt[3]{\left({x}^{3}{y}^{6}\right)}=x{y}^{2}$
$\sqrt[3]{40{x}^{4}{y}^{7}}=2x{y}^{2}\sqrt[3]{\left(5xy\right)}$
Therefore, the simplest form of the given expression $2x{y}^{2}\sqrt[3]{\left(5xy\right)}$

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