# Why is $\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$ $15\sqrt{5}$ and not $15\sqrt[4]{25}$?

I have an expression I am to simplify:

$$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$$

I arrived at $$15\sqrt[4]{25}$$. My textbook tells me that the answer is in fact $$15\sqrt{5}$$. Here is my thought process:

$$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$$ = $$\frac{15*\sqrt[4]{5}*\sqrt[4]{25}}{\sqrt[4]{5}}$$ =

(cancel out $$\sqrt[4]{5}$$ present in both numerator and denominator) leaving: $$15\sqrt[4]{25}$$

Where did I go wrong and how can I arrive at $$15\sqrt{5}$$?

• The fourth root is the square root of the square root ... Jan 3 '19 at 16:54

$$\frac{\sqrt[4] {125}}{\sqrt[4] 5} = \frac{\sqrt[4] {5^3}}{\sqrt[4] 5} = \sqrt[4]{5^2} = 5^{\frac{2}{4}} = 5^{\frac{1}{2}} = \sqrt 5$$
Even quicker, $$\sqrt[4]{25}$$ means $$\sqrt{\sqrt{25}}$$, which becomes $$\sqrt{5}$$.
Write $$\sqrt[4]{\frac{125}{5}}=\sqrt[4]{25}=\sqrt{5}$$
It turns out that $$\sqrt[4]{25}=\sqrt{5}$$. This is because $$25=5^2$$, so that $$\sqrt[4]{25}=\sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=\sqrt{5}.$$ So, you are correct, as is the book.