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.
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.