Question

Find the cube root of $135\sqrt{3}-87\sqrt{6}$.

Answer 1

Hint: This question can be solved easily if we can write $135\sqrt{3}-87\sqrt{6}$ in the form of ${{(a-b)}^{3}}$. Because it is easier to find the cube root, when it is represented as above.

Complete step-by-step answer:
We know that,
$\begin{align}
  & \Rightarrow {{(a-b)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab(a-b) \\
 & \Rightarrow {{(a-b)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}.......(i) \\
\end{align}$
Now, we have to analyse the question and check whether it can be written in the form equation(i). We will first write the question in the form of equation (i). Then we will find the cube root.

According to the question we have to find the cube root of $135\sqrt{3}-87\sqrt{6}$.
Now, we are going to simplify each term of the question
First, let us take $135\sqrt{3}$ and $87\sqrt{6}$.
 Here we can write $135\sqrt{3}$as,
$\Rightarrow 135\sqrt{3}=81\sqrt{3}+54\sqrt{3}.......(ii)$
Similarly, we can write $87\sqrt{6}$ as,
$\Rightarrow 87\sqrt{6}=81\sqrt{6}+6\sqrt{6}.......(iii)$
Now, we can rewrite the question as,
$\Rightarrow 135\sqrt{3}-87\sqrt{6}=(81\sqrt{3}+54\sqrt{3})-(81\sqrt{6}+6\sqrt{6})$
Opening the brackets,
$\Rightarrow 135\sqrt{3}-87\sqrt{6}=81\sqrt{3}+54\sqrt{3}-81\sqrt{6}-6\sqrt{6}$
Rearranging the above equation to get the form of equation(i) we have,
$\Rightarrow 135\sqrt{3}-87\sqrt{6}=81\sqrt{3}-6\sqrt{6}-81\sqrt{6}+54\sqrt{3}.......(iv)$
Here $81\sqrt{3}$ can be written as ${{\left( 3\sqrt{3} \right)}^{3}}$ . Similarly, $6\sqrt{6}$ can be written as ${{\left( \sqrt{6} \right)}^{3}}$.
Substituting this in equation (iv) we have,
$\Rightarrow 135\sqrt{3}-87\sqrt{6}={{\left( 3\sqrt{3} \right)}^{3}}-{{\left( \sqrt{6} \right)}^{3}}-81\sqrt{6}+54\sqrt{3}.......(v)$
Equation (v) we observe ${{a}^{3}}={{\left( 3\sqrt{3} \right)}^{3}}$ and ${{b}^{3}}={{\left( \sqrt{6} \right)}^{3}}$.
Hence, equation (v) can be written in the form of equation (i) as follows,
$\Rightarrow 135\sqrt{3}-87\sqrt{6}={{\left( 3\sqrt{3} \right)}^{3}}-{{\left( \sqrt{6} \right)}^{3}}-3\times 3\sqrt{3}\times \sqrt{6}\left( 3\sqrt{3}-\sqrt{6} \right).....(vi)$
Thus, from equation (vi) we observe that the question is rewritten in the form of equation (i).
Now, we can find the cube root.
Comparing equation (i) and equation (vi) we can simplify equation (vi) as,
$\Rightarrow 135\sqrt{3}-87\sqrt{6}={{\left( 3\sqrt{3}-\sqrt{6} \right)}^{3}}......(vii)$
Taking the cube root on both sides,
$\begin{align}
  & \Rightarrow \sqrt[3]{135\sqrt{3}-87\sqrt{6}}=\sqrt[3]{{{\left( 3\sqrt{3}-\sqrt{6} \right)}^{3}}} \\
 & \Rightarrow \sqrt[3]{135\sqrt{3}-87\sqrt{6}}={{\left( {{\left( 3\sqrt{3}-\sqrt{6} \right)}^{3}} \right)}^{\frac{1}{3}}} \\
 & \Rightarrow \sqrt[3]{135\sqrt{3}-87\sqrt{6}}=\left( 3\sqrt{3}-\sqrt{6} \right).......(viii) \\
\end{align}$
Hence, the cube root of $135\sqrt{3}-87\sqrt{6}$ is $3\sqrt{3}-\sqrt{6}$.

Note: The main idea is representing the problem in the form of,
\[\Rightarrow {{(a-b)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}\]. If this is correct then the student can easily simplify the above expression to obtain the cube root.