Answer

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**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.

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