We can rewrite the expression under the radical as
[tex]\dfrac{81}{16}a^8b^{12}c^{16}=\left(\dfrac32a^2b^3c^4\right)^4[/tex]
then taking the fourth root, we get
[tex]\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|[/tex]
Why the absolute value? It's for the same reason that
[tex]\sqrt{x^2}=|x|[/tex]
since both [tex](-x)^2[/tex] and [tex]x^2[/tex] return the same number [tex]x^2[/tex], and [tex]|x|[/tex] captures both possibilities. From here, we have
[tex]\left|\dfrac32a^2b^3c^4\right|=\left|\dfrac32\right|\left|a^2\right|\left|b^3\right|\left|c^4\right|[/tex]
The absolute values disappear on all but the [tex]b[/tex] term because all of [tex]\dfrac32[/tex], [tex]a^2[/tex] and [tex]c^4[/tex] are positive, while [tex]b^3[/tex] could potentially be negative. So we end up with
[tex]\dfrac32a^2\left|b^3\right|c^4=\dfrac32a^2|b|^3c^4[/tex]
