simplify. remove all perfect squares from inside the square root. $sqrt{112a^{6}}=$

Explanation:

Step1: Factor 112 and $a^6$

Decompose 112 into prime - factors: $112 = 16\times7$, and $a^6=(a^3)^2$. So $\sqrt{112a^6}=\sqrt{16\times7\times(a^3)^2}$.

Step2: Use the square - root property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$

$\sqrt{16\times7\times(a^3)^2}=\sqrt{16}\cdot\sqrt{(a^3)^2}\cdot\sqrt{7}$.

Step3: Simplify each square - root

$\sqrt{16} = 4$, $\sqrt{(a^3)^2}=|a^3|$. So $\sqrt{16}\cdot\sqrt{(a^3)^2}\cdot\sqrt{7}=4|a^3|\sqrt{7}$.

Answer:

$4|a^3|\sqrt{7}$