Question

Simplify the expression: $\dfrac{6}{2\sqrt{3}-\sqrt{6}}+\dfrac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}-\dfrac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}$

Answer 1

Hint: We solve this problem by using the rationalising concept. We rationalise the denominator of each term by multiplying the numerator and denominator with the specific term to change the denominator to the whole number. Then we get the irrational numbers only in numerator which will be easy to find the addition or subtraction.
We use the standard formula of algebra that is,
$\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$

Complete step-by-step solution:
We are asked to simplify the expression given as $\dfrac{6}{2\sqrt{3}-\sqrt{6}}+\dfrac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}-\dfrac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}$
Let us assume that the required value as,
$\Rightarrow A=\dfrac{6}{2\sqrt{3}-\sqrt{6}}+\dfrac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}-\dfrac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}..............equation(i)$
Now, let us take each term separately and rationalise the denominator.
Let us take the first term of given expression as,
$\Rightarrow x=\dfrac{6}{2\sqrt{3}-\sqrt{6}}$
Here, we can see that the denominator is $2\sqrt{3}-\sqrt{6}$
So, let us multiply the numerator and denominator with $2\sqrt{3}+\sqrt{6}$ to rationalise the denominator then we get,
$\Rightarrow x=\dfrac{6\left( 2\sqrt{3}+\sqrt{6} \right)}{\left( 2\sqrt{3}-\sqrt{6} \right)\left( 2\sqrt{3}+\sqrt{6} \right)}$
We know that the standard formula of algebra that is,
$\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
By using the standard result of algebra in above equation then we get,
\[\begin{align}
  & \Rightarrow x=\dfrac{6\left( 2\sqrt{3}+\sqrt{6} \right)}{{{\left( 2\sqrt{3} \right)}^{2}}-{{\left( \sqrt{6} \right)}^{2}}} \\
 & \Rightarrow x=\dfrac{6\left( 2\sqrt{3}+\sqrt{6} \right)}{12-6} \\
 & \Rightarrow x=\left( 2\sqrt{3}+\sqrt{6} \right) \\
\end{align}\]
Let us take the second term of given expression as,
$\Rightarrow y=\dfrac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}$
Here, we can see that the denominator is $\sqrt{3}+\sqrt{2}$
So, let us multiply the numerator and denominator with $\sqrt{3}-\sqrt{2}$ to rationalise the denominator then we get,
$\Rightarrow y=\dfrac{\sqrt{6}\left( \sqrt{3}-\sqrt{2} \right)}{\left( \sqrt{3}+\sqrt{2} \right)\left( \sqrt{3}-\sqrt{2} \right)}$
By using the standard result of algebra in above equation then we get,
\[\begin{align}
  & \Rightarrow y=\dfrac{\sqrt{6}\left( \sqrt{3}-\sqrt{2} \right)}{{{\left( \sqrt{3} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}} \\
 & \Rightarrow y=\dfrac{\sqrt{6}\left( \sqrt{3}-\sqrt{2} \right)}{3-2} \\
\end{align}\]
Here, we can see that the term $\sqrt{6}$ can be represented as $\sqrt{3}\times \sqrt{2}$ then we get,
$\begin{align}
  & \Rightarrow y=\left( \sqrt{3}\times \sqrt{2}\times \sqrt{3} \right)-\left( \sqrt{3}\times \sqrt{2}\times \sqrt{2} \right) \\
 & \Rightarrow y=3\sqrt{2}-2\sqrt{3} \\
\end{align}$
Let us take the third term of given expression as,
$\Rightarrow z=\dfrac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}$
Here, we can see that the denominator is $\sqrt{6}-\sqrt{2}$
So, let us multiply the numerator and denominator with $\sqrt{6}+\sqrt{2}$ to rationalise the denominator then we get,
$\Rightarrow z=\dfrac{4\sqrt{3}\left( \sqrt{6}+\sqrt{2} \right)}{\left( \sqrt{6}-\sqrt{2} \right)\left( \sqrt{6}+\sqrt{2} \right)}$
By using the standard result of algebra in above equation then we get,
\[\begin{align}
  & \Rightarrow z=\dfrac{4\sqrt{3}\left( \sqrt{6}+\sqrt{2} \right)}{{{\left( \sqrt{6} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}} \\
 & \Rightarrow z=\dfrac{4\sqrt{3}\left( \sqrt{6}+\sqrt{2} \right)}{6-2} \\
 & \Rightarrow z=\sqrt{3}\left( \sqrt{6}+\sqrt{2} \right) \\
\end{align}\]
Here, we can see that the term $\sqrt{6}$ can be represented as $\sqrt{3}\times \sqrt{2}$ then we get,
$\begin{align}
  & \Rightarrow z=\left( \sqrt{3}\times \sqrt{2}\times \sqrt{3} \right)+\left( \sqrt{3}\times \sqrt{2} \right) \\
 & \Rightarrow z=3\sqrt{2}+\sqrt{6} \\
\end{align}$
Now, let us take the equation (i) and substitute the required values then we get,
$\begin{align}
  & \Rightarrow A=x+y-z \\
 & \Rightarrow A=\left( 2\sqrt{3}+\sqrt{6} \right)+\left( 3\sqrt{2}-2\sqrt{3} \right)-\left( 3\sqrt{2}+\sqrt{6} \right) \\
 & \Rightarrow A=0 \\
\end{align}$
Therefore, we can conclude the value of given expression is 0 that is,
$\therefore \dfrac{6}{2\sqrt{3}-\sqrt{6}}+\dfrac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}-\dfrac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}=0$

Note: The main mistake that can be done in this question is due to calculation mistake only. We need to use the correct number for rationalising the denominator of each term. If the denominator is of the form $a+b$ then we need to multiply the numerator and denominator with $a-b$ to rationalise it. But there is no need to rationalise the numerator because adding the radical terms in the numerator is what we want.