Question

If \[f\left( x \right)={{x}^{3}}-\dfrac{1}{{{x}^{3}}}\] then show that \[f\left( x \right)+f\left( \dfrac{1}{x} \right)=0\] also find the number of zeros of
\[f\left( x \right)\]

Answer 1

Hint: We simply solve this problem by finding the value of \[f\left( \dfrac{1}{x} \right)\] simply by replacing \['x'\] with \['\dfrac{1}{x}'\] to prove the required result.
Then we find the number of zeros of \[f\left( x \right)\] by making the given polynomial to zero that is we take
\[\Rightarrow f\left( x \right)=0\]
Then we find the roots of the above equation.

Complete step by step answer:
We are given that
\[f\left( x \right)={{x}^{3}}-\dfrac{1}{{{x}^{3}}}......equation(i)\]
Now, let us replace \['x'\] with \['\dfrac{1}{x}'\] then we get
\[\begin{align}
  & \Rightarrow f\left( \dfrac{1}{x} \right)={{\left( \dfrac{1}{x} \right)}^{3}}-\dfrac{1}{{{\left( \dfrac{1}{x} \right)}^{3}}} \\
 & \Rightarrow f\left( \dfrac{1}{x} \right)=\dfrac{1}{{{x}^{3}}}-{{x}^{3}}........equation(ii) \\
\end{align}\]
Now, by adding the equation (i) and equation (ii) we get
\[\begin{align}
  & \Rightarrow f\left( x \right)+f\left( \dfrac{1}{x} \right)=\left( {{x}^{3}}-\dfrac{1}{{{x}^{3}}} \right)+\left( \dfrac{1}{{{x}^{3}}}-{{x}^{3}} \right) \\
 & \Rightarrow f\left( x \right)+f\left( \dfrac{1}{x} \right)=0 \\
\end{align}\]
Hence the required result has been proved.
Now let us find the zeros of \[f\left( x \right)\]
We know that the zeros of polynomial can be found by taking the polynomial to zero that is
\[\Rightarrow f\left( x \right)=0\]
By substituting the value of \[f\left( x \right)\] we get
\[\Rightarrow {{x}^{3}}-\dfrac{1}{{{x}^{3}}}=0\]
By adding the terms and cross multiplying we get
\[\begin{align}
  & \Rightarrow {{x}^{6}}-1=0 \\
 & \Rightarrow {{\left( {{x}^{3}} \right)}^{2}}-{{1}^{2}}=0 \\
\end{align}\]
We know that the standard formula of algebra that is
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
By using this formula we get
\[\Rightarrow \left( {{x}^{3}}+1 \right)\left( {{x}^{3}}-1 \right)=0\]
We know that if \[a\times b=0\] then either of \[a,b\] will be zero.
By using the above result let us take the first term that is
\[\begin{align}
  & \Rightarrow {{x}^{3}}+1=0 \\
 & \Rightarrow x=-1 \\
\end{align}\]
Here, we can say that there are 3 roots all are equal to ‘-1’
Now, let us take the second term that is
\[\begin{align}
  & \Rightarrow {{x}^{3}}-1=0 \\
 & \Rightarrow x=1 \\
\end{align}\]
Here, we can say that there are 3 roots all are equal to ‘1’

Therefore, the number of roots of \[f\left( x \right)=0\] are 6 in which 3 roots are equal to ‘1’ and other 3 roots are equal to ‘-1’.

Note: Students may make mistakes in the second part that is the number of zeros.
Here we have
\[\begin{align}
  & \Rightarrow {{x}^{3}}+1=0 \\
 & \Rightarrow x=-1 \\
\end{align}\]
Here the number of roots is 3 not 1 because the equation is a cubic equation in which it has 3 roots but they are equal.
Even though the roots are equal, we need to give the number of roots as 3.
This point needs to be taken care of.