Question

If $a=x+\dfrac{1}{x},\ then\ {{x}^{3}}+{{x}^{-3}}=\_\_\_\_\_$.
A. ${{a}^{3}}+3a$
B. ${{a}^{3}}-3a$
C. ${{a}^{3}}+3$
D. ${{a}^{3}}-3$

Answer 1

- Hint: Cube the given equation on both sides. Rearrange the equation to get the value of ${{x}^{3}}+\dfrac{1}{{{x}^{3}}}$
 i.e. ${{x}^{3}}+{{x}^{-3}}$. Then put the value of $x+\dfrac{1}{x}$ equal to a in the expression. Finally, you will get the
value of ${{x}^{3}}+{{x}^{-3}}$ in terms of a.

Complete step-by-step solution -

Given equation,
$\Rightarrow a=x+\dfrac{1}{x}$
On cubing the above equation both sides, we get,
$\Rightarrow {{\left( a \right)}^{3}}={{\left( x+\dfrac{1}{x} \right)}^{3}}..........\left( 1 \right)$
We know that, ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}$
Using this formula, expanding equation (1), we get,
$\begin{align}
  & \Rightarrow {{a}^{3}}={{\left( x \right)}^{3}}+{{\left( \dfrac{1}{x} \right)}^{3}}+3{{x}^{2}}\left( \dfrac{1}{x} \right)+3\left( x \right){{\left( \dfrac{1}{x} \right)}^{2}} \\
 & \Rightarrow {{a}^{3}}={{x}^{3}}+\dfrac{1}{{{x}^{3}}}+3x+\dfrac{3}{x} \\
\end{align}$
Taking 3 common from the last two terms in the right hand side, we get,
$\Rightarrow {{a}^{3}}={{x}^{3}}+\dfrac{1}{{{x}^{3}}}+3\left( x+\dfrac{1}{x} \right)........\left( 2 \right)$
According to the question, it is given that,
$a=x+\dfrac{1}{x}$
Now, putting the value of $x+\dfrac{1}{x}$ equal to a in equation (2), we will get,
\[\begin{align}
  & \Rightarrow {{a}^{3}}={{x}^{3}}+\dfrac{1}{{{x}^{3}}}+3a \\
 & \Rightarrow {{a}^{3}}={{x}^{3}}+{{x}^{-3}}+3a\ \ \ \ \ \ \ \ \left[ \text{since, }\dfrac{1}{{{x}^{n}}}={{x}^{-n}} \right] \\
 & \Rightarrow {{a}^{3}}-3a={{x}^{3}}+{{x}^{-3}} \\
 & Hence,\ {{x}^{3}}+{{x}^{-3}}={{a}^{3}}-3a \\
\end{align}\]

Note: We can also do this question in the given way,
Since,
$\begin{align}
  & {{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy \\
 & \therefore {{\left( x+\dfrac{1}{x} \right)}^{2}}={{x}^{2}}+\dfrac{1}{{{x}^{2}}}+2\times x\times \dfrac{1}{x} \\
 & \therefore {{x}^{2}}+\dfrac{1}{{{x}^{2}}}={{\left( x+\dfrac{1}{x} \right)}^{2}}-2 \\
 & \therefore {{x}^{2}}+\dfrac{1}{{{x}^{2}}}=\left( {{a}^{2}}-2 \right)\ \left( \because x+\dfrac{1}{x}=a \right) \\
\end{align}$
Now, we know that,
$\begin{align}
  & {{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right) \\
 & \therefore {{x}^{3}}+\dfrac{1}{{{x}^{3}}}=\left( x+\dfrac{1}{x} \right)\left( {{x}^{2}}-x\times \dfrac{1}{x}+\dfrac{1}{{{x}^{2}}} \right) \\
 & \Rightarrow {{x}^{3}}+{{x}^{-3}}=a\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}}-1 \right)\ \ \left( \because x+\dfrac{1}{x}=a \right) \\
 & \Rightarrow {{x}^{3}}+{{x}^{-3}}=a\left( {{a}^{2}}-2-1 \right)\ \ \ \ \ \ \ \ \ \ \left\{ \because {{x}^{2}}+\dfrac{1}{{{x}^{2}}}={{a}^{2}}-2\left( obtained\ above \right) \right\} \\
 & \Rightarrow {{x}^{3}}+{{x}^{-3}}=a\left( {{a}^{2}}-3 \right) \\
 & \Rightarrow {{x}^{3}}+{{x}^{-3}}={{a}^{3}}-3a \\
\end{align}$
Hence, option B is correct.