Question

If $x={{2}^{\dfrac{1}{3}}}-{{2}^{\dfrac{-1}{3}}}$ then $2{{x}^{3}}+6x$ is equal to
A) $0$
B) $3$
C) $\dfrac{3}{2}$
D) $6$

Answer 1

Hint: To solve the above question we will simply use mathematical operations like addition, subtraction, multiplication and division. In the given the power of $2$ is in fraction form. To get the value of $2{{x}^{3}}+6x$, first we have to know the value of x. since if we see carefully the value of x is given which is $x={{2}^{\dfrac{1}{3}}}-{{2}^{\dfrac{-1}{3}}}$, but direct putting this value in the given equation it may give wrong answer. So to make calculation easy first we will simplify the value of x.

Complete step by step solution:
The given value of x is:
$x={{2}^{\dfrac{1}{3}}}-{{2}^{\dfrac{-1}{3}}}$
Here ${{2}^{\dfrac{-1}{3}}}$ can be written as $\dfrac{1}{{{2}^{\dfrac{1}{3}}}}$ by using the rule${{a}^{-b}}=\dfrac{1}{{{a}^{b}}}$
So by putting this in the above given expression we get
$\Rightarrow x={{2}^{\dfrac{1}{3}}}-\dfrac{1}{{{2}^{\dfrac{1}{3}}}}$
Now to more simplify, we will take LCM of the above expression, then we get
$\Rightarrow x=\dfrac{{{2}^{\dfrac{2}{3}}}-1}{{{2}^{\dfrac{1}{3}}}}$
Now we will put this value of x in the given equation $2{{x}^{3}}+6x$, then we get
$\begin{align}
  & 2{{\left( \dfrac{{{2}^{\dfrac{2}{3}}}-1}{{{2}^{\dfrac{1}{3}}}} \right)}^{3}}+6\left( \dfrac{{{2}^{\dfrac{2}{3}}}-1}{{{2}^{\dfrac{1}{3}}}} \right) \\
 & \\
\end{align}$
Now taking $\dfrac{{{2}^{\dfrac{2}{3}}}-1}{{{2}^{\dfrac{1}{3}}}}$ common from the above expression, then we get
 $\begin{align}
  & = \dfrac{{{2}^{\dfrac{2}{3}}}-1}{{{2}^{\dfrac{1}{3}}}}\left[ 2{{\left( \dfrac{{{2}^{\dfrac{2}{3}}}-1}{{{2}^{\dfrac{1}{3}}}} \right)}^{2}}+6 \right] \\
 & \dfrac{{{2}^{\dfrac{2}{3}}}-1}{{{2}^{\dfrac{1}{3}}}}\left[ \dfrac{2}{{{2}^{\dfrac{2}{3}}}}{{\left( {{2}^{\dfrac{2}{3}}}-1 \right)}^{2}}+6 \right] \\
\end{align}$
Now we know ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ , so when we open ${{\left( {{2}^{\dfrac{2}{3}}}-1 \right)}^{2}}$we get
$\begin{align}
  & = \dfrac{{{2}^{\dfrac{2}{3}}}-1}{{{2}^{\dfrac{1}{3}}}}\left[ \dfrac{2}{{{2}^{\dfrac{2}{3}}}}\left( {{2}^{\dfrac{4}{3}}}+1-2\times {{2}^{\dfrac{2}{3}}} \right)+6 \right] \\
 & \dfrac{{{2}^{\dfrac{2}{3}}}-1}{{{2}^{\dfrac{1}{3}}}}\left[ {{2}^{\dfrac{1}{3}}}\left( {{2}^{\dfrac{4}{3}}}+1-{{2}^{\dfrac{5}{3}}} \right)+6 \right] \\
 & = \dfrac{{{2}^{\dfrac{2}{3}}}-1}{{{2}^{\dfrac{1}{3}}}}\left[ {{2}^{\dfrac{5}{3}}}+{{2}^{\dfrac{1}{3}}}-{{2}^{2}}+6 \right] \\
\end{align}$
Now by more simplifying the above, we get
$\begin{align}
  & = \dfrac{{{2}^{\dfrac{2}{3}}}-1}{{{2}^{\dfrac{1}{3}}}}\left[ {{2}^{\dfrac{5}{3}}}+{{2}^{\dfrac{1}{3}}}-4+6 \right] \\
 & = \dfrac{{{2}^{\dfrac{2}{3}}}-1}{{{2}^{\dfrac{1}{3}}}}\left[ {{2}^{\dfrac{5}{3}}}+{{2}^{\dfrac{1}{3}}}+2 \right] \\
 & = \left( {{2}^{\dfrac{2}{3}}}-1 \right)\left[ \dfrac{{{2}^{\dfrac{5}{3}}}}{{{2}^{\dfrac{1}{3}}}}+\dfrac{{{2}^{\dfrac{1}{3}}}}{{{2}^{\dfrac{1}{3}}}}+\dfrac{2}{{{2}^{\dfrac{1}{3}}}} \right] \\
\end{align}$
We know $\dfrac{{{a}^{b}}}{{{a}^{c}}}={{a}^{b-c}}$so apply this on the above expression, we get
$\begin{align}
  & = \left( {{2}^{\dfrac{2}{3}}}-1 \right)\left[ {{2}^{\dfrac{5-1}{3}}}+{{2}^{\dfrac{1-1}{3}}}+{{2}^{\dfrac{3-1}{3}}} \right] \\
 & = \left( {{2}^{\dfrac{2}{3}}}-1 \right)\left[ {{2}^{\dfrac{4}{3}}}+1+{{2}^{\dfrac{2}{3}}} \right] \\
\end{align}$
Now multiplying the $\left( {{2}^{\dfrac{2}{3}}}-1 \right)$inside the expression, we will get some similar term which get cancel out,
$\begin{align}
  & = {{2}^{\dfrac{2}{3}+\dfrac{4}{3}}}+{{2}^{\dfrac{2}{3}}}+{{2}^{\dfrac{2}{3}+\dfrac{2}{3}}}-{{2}^{\dfrac{4}{3}}}-1-{{2}^{\dfrac{2}{3}}} \\
 & = {{2}^{2}}-1=4-1 \\
 & = 3 \\
\end{align}$
Hence the value of the given equation $2{{x}^{3}}+6x$ is $3$.

Note: It is not hard to solve these types question but due to lengthy process students make mistakes. It's calculation part take time to solve, so I would to suggest whenever you have these types of question first try to make the value of any variable in a simple way so that you can easily get the value. In this question we used properties of mathematics such as ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, ${{a}^{-b}}=\dfrac{1}{{{a}^{b}}}$,$\dfrac{{{a}^{b}}}{{{a}^{c}}}={{a}^{b-c}}$and${{a}^{b}}\cdot {{a}^{c}}={{a}^{b+c}}$ .