Question

If ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}=51$, find the value of ${{x}^{3}}-\dfrac{1}{{{x}^{3}}}$.

Answer 1

Hint: We will first start by using the fact ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$. Then we will use the given data that ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}=51$ to find the value of $x-\dfrac{1}{x}$ by using the fact that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-ab$. Then we will use this formula to find the value of $x-\dfrac{1}{x}$ and then use this value to find the value of ${{x}^{3}}-\dfrac{1}{{{x}^{3}}}$.

Complete step-by-step answer:
Now, we have been given that ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}=51$.
We know the algebraic identity that,
${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
So, using this identity we have,
$\begin{align}
  & {{\left( x-\dfrac{1}{x} \right)}^{2}}={{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2\left( x \right)\left( \dfrac{1}{x} \right) \\
 & {{\left( x-\dfrac{1}{x} \right)}^{2}}={{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2 \\
\end{align}$
Now, we know that the value of ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$ has been given to us as 51. So, we have,
\[\begin{align}
  & {{\left( x-\dfrac{1}{x} \right)}^{2}}=51-2 \\
 & {{\left( x-\dfrac{1}{x} \right)}^{2}}=49 \\
 & x-\dfrac{1}{x}=\sqrt{49} \\
 & x-\dfrac{1}{x}=7..........\left( 1 \right) \\
\end{align}\]
Now, we have to find the value of ${{x}^{3}}-\dfrac{1}{{{x}^{3}}}$. We know the fact that,
${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$
So, we have the value of,
${{x}^{3}}-\dfrac{1}{{{x}^{3}}}=\left( x-\dfrac{1}{x} \right)\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}}+x\times \dfrac{1}{x} \right)$
Now, we will substitute the value of ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$ as given to us in the question and $x-\dfrac{1}{x}$ from (1). So, we have,
$\begin{align}
  & {{x}^{3}}-\dfrac{1}{{{x}^{3}}}=7\left( 51+1 \right) \\
 & =7\times 52 \\
 & =364 \\
\end{align}$
Hence, the value of ${{x}^{3}}-\dfrac{1}{{{x}^{3}}}$ is 364.

Note: To solve this question we have used the algebraic identity that ${{\left( a-b \right)}^{2}} = {{a}^{2}}+{{b}^{2}}-ab$. It is an important trick to remember as in the case of $a=x,b=\dfrac{1}{x}$ or vice versa. We have $ab=1$, hence we can easily find the value of $x-\dfrac{1}{x}\ or\ x+\dfrac{1}{x}$ given the value of ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$.