Question

If ${{x}^{3}}-\dfrac{1}{{{x}^{3}}}=14$, then find the value of $x-\dfrac{1}{x}$.

Answer 1

Hint:Use the formula: ${{a}^{3}}-{{b}^{3}}={{\left( a-b \right)}^{3}}+3ab\left( a-b \right)$ and substitute, $x=a$ and $\dfrac{1}{x}=b$. An equation will be formed with the variable given as, $x-\dfrac{1}{x}$. The next step is to assume $x-\dfrac{1}{x}$ equal to ‘m’. A cubic equation will be formed. Find one root of the cubic equation by hit and trial method and the other two roots by the discriminant method, given by: $\text{Discriminant}={{B}^{2}}-4AC$. The value of ‘m’ obtained will be the answer.

Complete step-by-step answer:
We have been given, ${{x}^{3}}-\dfrac{1}{{{x}^{3}}}=14$. This is of the form: ${{a}^{3}}-{{b}^{3}}=c$.
Now, substituting $a=x,b=\dfrac{1}{x}\text{ and }c=14$, we get, ${{x}^{3}}-\dfrac{1}{{{x}^{3}}}=14$.
Applying the formula: ${{a}^{3}}-{{b}^{3}}={{\left( a-b \right)}^{3}}+3ab\left( a-b \right)$, we get,
${{x}^{3}}-\dfrac{1}{{{x}^{3}}}={{\left( x-\dfrac{1}{x} \right)}^{3}}+3x\times \dfrac{1}{x}\left( x-\dfrac{1}{x} \right)$
Now, substituting the value, ${{x}^{3}}-\dfrac{1}{{{x}^{3}}}=14$, we have,
$\begin{align}
  & 14={{\left( x-\dfrac{1}{x} \right)}^{3}}+3x\times \dfrac{1}{x}\left( x-\dfrac{1}{x} \right) \\
 & \Rightarrow {{\left( x-\dfrac{1}{x} \right)}^{3}}+3x\times \dfrac{1}{x}\left( x-\dfrac{1}{x} \right)=14 \\
 & \Rightarrow {{\left( x-\dfrac{1}{x} \right)}^{3}}+3\left( x-\dfrac{1}{x} \right)=14 \\
\end{align}$
Assuming, $x-\dfrac{1}{x}=m$, we get,
$\begin{align}
  & {{m}^{3}}+3m=14 \\
 & {{m}^{3}}+3m-14=0 \\
\end{align}$
Clearly, we can see that this is a cubic equation in ‘m’. Therefore, we have to find its one root by hit and trial method.
One can see that if we will substitute (m = 2) in the above cubic equation, we get,
$\begin{align}
  & {{2}^{3}}+3\times 2-14 \\
 & =8+6-14 \\
 & =14-14 \\
 & =0 \\
\end{align}$
Hence, m = 2 satisfies the cubic equation, therefore, $(m-2)$ is a factor of the given equation. Therefore, the equation can be written as: $\left( m-2 \right)\left( {{m}^{2}}+2m+7 \right)=0$.
Now, if we will substitute $\left( {{m}^{2}}+2m+7 \right)=0$, then we have to find its root.
Let us find the root by discriminant method.
First, we have to find the discriminant of the equation, $\left( {{m}^{2}}+2m+7 \right)$. Therefore,
$\text{Discriminant}={{B}^{2}}-4AC$, where B = 2, A=1 and C= 7.
Now, for the equation to have real roots, the value of the discriminant must be greater than or equal to zero. So, let us see the value of discriminant.
$\Rightarrow \text{Discriminant}={{2}^{2}}-4\times 1\times 7=4-28=-24$.
Clearly, the discriminant is a negative value. Hence, real roots of $\left( {{m}^{2}}+2m+7 \right)=0$ does not exist.
Therefore, there is only one root of the cubic equation: ${{m}^{3}}+3m-14=0$ and that is m = 2.
Therefore, $x-\dfrac{1}{x}=2$.

Note: One may note that a cubic equation can have either one real root with two imaginary roots or it can have three real roots. In the above question, we have a cubic equation that has one real root with two imaginary roots. Hence, there is only one value of $x-\dfrac{1}{x}$ because we do not have to consider the imaginary values.