Question

Simplify

Answer 1

Answer:

$\huge\boxed{\sqrt[4]{16a^{-12}}=2a^{-3}=\dfrac{2}{a^3}}$

Step-by-step explanation:

$16=2^4\\\\a^{-12}=a^{(-3)(4)}=\left(a^{-3}\right)^4\qquad\text{used}\ (a^n)^m=a^{nm}\\\\\sqrt[4]{16a^{-12}}=\bigg(16a^{-12}\bigg)^\frac{1}{4}\qquad\text{used}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\=\bigg(2^4(a^{-3})^4\bigg)^\frac{1}{4}\qquad\text{use}\ (ab)^n=a^nb^n\\\\=\bigg(2^4\bigg)^\frac{1}{4}\bigg[(a^{-3})^4\bigg]^\frac{1}{4}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=2^{(4)(\frac{1}{4})}(a^{-3})^{(4)(\frac{1}{4})}=2^1(a^{-3})^1=2a^{-3}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}$

$=2\left(\dfrac{1}{a^3}\right)=\dfrac{2}{a^3}$