Question

Could someone help me, please?

Answer 1

We can rewrite the expression under the radical as

$\dfrac{81}{16}a^8b^{12}c^{16}=\left(\dfrac32a^2b^3c^4\right)^4$

then taking the fourth root, we get

$\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|$

Why the absolute value? It's for the same reason that

$\sqrt{x^2}=|x|$

since both $(-x)^2$ and $x^2$ return the same number $x^2$, and $|x|$ captures both possibilities. From here, we have

$\left|\dfrac32a^2b^3c^4\right|=\left|\dfrac32\right|\left|a^2\right|\left|b^3\right|\left|c^4\right|$

The absolute values disappear on all but the $b$ term because all of $\dfrac32$, $a^2$ and $c^4$ are positive, while $b^3$ could potentially be negative. So we end up with

$\dfrac32a^2\left|b^3\right|c^4=\dfrac32a^2|b|^3c^4$