Simplify by factoring the cube root of 7/y, assuming y is a positive real number.

Step-by-step solution

1. 1

   Split the radical over the fraction. The quotient rule for radicals gives $\sqrt[3]{\frac{7}{y}}=\frac{\sqrt[3]{7}}{\sqrt[3]{y}}.$ This is not yet a simplified form, because standard form requires no radical in the denominator.

2. 2

   Decide what makes the denominator a perfect cube. For a cube root the denominator needs a factor of $y^{3}$ inside the radical, and it currently has only $y^{1}$. Multiplying by $\sqrt[3]{y^{2}}$ supplies the missing two factors, since $\sqrt[3]{y}\cdot\sqrt[3]{y^{2}}=\sqrt[3]{y^{3}}=y$. This is why the cube-root case uses $y^{2}$ where a square root would use $y$.

3. 3

   Multiply top and bottom by that factor. $\frac{\sqrt[3]{7}}{\sqrt[3]{y}}\cdot\frac{\sqrt[3]{y^{2}}}{\sqrt[3]{y^{2}}}=\frac{\sqrt[3]{7y^{2}}}{\sqrt[3]{y^{3}}}.$

4. 4

   Simplify the denominator. Because $y>0$, $\sqrt[3]{y^{3}}=y$, so $\sqrt[3]{\frac{7}{y}}=\frac{\sqrt[3]{7y^{2}}}{y}.$

5. 5

   Confirm the form is fully simplified. The radicand $7y^{2}$ contains no factor that is a perfect cube, the index cannot be reduced, and no radical remains in the denominator, so this is the standard simplified answer.