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**HallsofIvy** $\displaystyle 2^4= 16$. Since, in the original problem, there was a "2" in the denominator, they multiplied by 8 to get a "perfect fourth power". Similarly, $\displaystyle b^3(b)= b^4$ and $\displaystyle c^2(c^2)= c^4$. Multiplying the denominator by $\displaystyle 8bc^2$ makes the denominator $\displaystyle 16b^4c^4= (2bc)^4$ so that $\displaystyle \sqrt[4]{(2bc)^4}= 2bc$.

Of course, if you multiply by $\displaystyle 8bc^2$ in the denominator you must do the same in the numerator.