The simplest form of the expression, 40x4y73

Given: $\sqrt[3]{40x^4{y}^7}$
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:
$\sqrt[3]{40x^4{y}^7}=\sqrt[3]{5\times 8\times x^3\times x\times y^6\times y}$
Grouping the perfect cubes, we get:
$\sqrt[3]{40x^4{y}^7}=\sqrt[3]{(8x^3{y}^6)(5xy)}$
Factoring into two radicals, we get:
$\sqrt[3]{40x^4{y}^7}=\sqrt[3]{(8x^3{y}^6)}\times \sqrt[3]{(5xy)}$
$\sqrt[3]{8}=2$ and $\sqrt[3]{(x^3{y}^6)}=x{y}^2$
$\sqrt[3]{40x^4{y}^7}=2x{y}^2\sqrt[3]{(5xy)}$
Therefore, the simplest form of the given expression $2x{y}^2\sqrt[3]{(5xy)}$