Question

Solve the expression, $\sqrt[3]{{{g}^{6}}{{h}^{3}}-27g{{h}^{3}}}$ given that $g,h>0$.
A. ${{g}^{2}}h-3gh\sqrt[2]{g}$
B. $gh\sqrt[3]{{{g}^{3}}-27g}$
C. $\dfrac{1}{3}gh\sqrt[3]{{{g}^{2}}-27}$
D. ${{g}^{2}}h-3\sqrt[3]{g}$

Answer 1

For solving this question you should know about the solution of general powered expression. Generally we solve these types of more powered equations by reducing the power of the variables as much as possible. And when these are reduced completely, then we solve that for the variable. In this problem we also do the same. We will first write the expression in the proper power way and then we solve it by reducing the power of the variables and then simplifying it.

Complete step-by-step solution:
According to the question it is asked to us to solve the expression, which is given as $\sqrt[3]{{{g}^{6}}{{h}^{3}}-27g{{h}^{3}}}$ given that $g,h>0$. For solving the problem first we will reduce the powers of the variables. As we know that the expression is given as: $\sqrt[3]{{{g}^{6}}{{h}^{3}}-27g{{h}^{3}}},g,h>0$. We can also write this equation in simple form as:
${{\left( {{g}^{6}}{{h}^{3}}-27g{{h}^{3}} \right)}^{\dfrac{1}{3}}}$
And if we take $g$ and $h$ common from the whole expression for reducing the power, then it can be written as:
${{\left( {{g}^{3}}.{{h}^{3}}\left( {{g}^{3}}-27g \right) \right)}^{\dfrac{1}{3}}}$
Now if we solve to the powers of this, then it will be:
$\begin{aligned} & {{g}^{\dfrac{3}{3}}}.{{h}^{\dfrac{3}{3}}}{{\left( {{g}^{3}}-27g \right)}^{\dfrac{1}{3}}} \\\ & =g.h{{\left( {{g}^{3}}-27g \right)}^{\dfrac{1}{3}}} \\\ & =g.h\sqrt[3]{{{g}^{3}}-27g} \\\ \end{aligned}$
Hence the correct answer is option B.

Note: While solving these type of questions, you have to keep in mind that the value of square root is $\dfrac{1}{2}$ and the cube root is $\dfrac{1}{3}$. It means we can write these as the power of that expression or variable as $\dfrac{1}{2}$ and $\dfrac{1}{3}$ respectively. And then we proceed forward.