# How do you simplify 8sqrt(11b^4) - sqrt(99b^4)?

May 18, 2015

By exponential definition, we know that $\sqrt[n]{{a}^{m}} = {a}^{\frac{m}{n}}$

In this case, we can do the same for $b$, factoring it out:

$8 {b}^{2} \sqrt{11} - {b}^{2} \sqrt{99}$
However, we can see that $99 = 11 \cdot 9$, so

$8 {b}^{2} \sqrt{11} - {b}^{2} \sqrt{11 \cdot 9}$

However, $9$ is a square number. Thus, we can take its root out of the square root.

$8 {b}^{2} \sqrt{11} - 3 {b}^{2} \sqrt{11}$

However, as both factors are multiplying ${b}^{2} \sqrt{11}$, we can factor it in function of such elements:

${b}^{2} \sqrt{11} \left(8 - 3\right)$