Question

If \[\alpha \], \[\beta \]and \[\gamma \] are the roots of the equation \[{{x}^{3}}-7x+7=0\], then \[\left( \dfrac{1}{{{\alpha }^{4}}} \right)+\left( \dfrac{1}{{{\beta }^{4}}} \right)+\left( \dfrac{1}{{{\gamma }^{4}}} \right)=\]

Answer 1

Hint: To find the value of the equation which is given in the question. First, we have to obtain the value of equations which are formed by combining the roots. The value of the sum of the roots, the product of the roots, and the sum of the product of the roots will be obtained from the equation given in the question and then the question is solved further.

Complete step-by-step answer:
An equation can be represented in the form of a linear polynomial where the highest degree of x is \[1\]. An equation can also be represented in the form of a binomial, where the highest degree of x is \[2\]and an equation can also be represented in the form of a cubic polynomial, where the highest degree of x is \[3\].
So in the above question, we have a cubic equation. So if a cubic equation has \[\alpha \], \[\beta \], and \[\gamma \]as the roots of the equation then that equation can be written as shown below.
\[{{x}^{3}}-(\alpha +\beta +\gamma ){{x}^{2}}+(\alpha \beta +\beta \gamma +\gamma \alpha )x+\alpha \beta \gamma =0\]
Where \[\alpha +\beta +\gamma \] is the coefficient of \[{{x}^{2}}\]. \[\alpha \beta +\beta \gamma +\alpha \gamma \] is the coefficient of x and \[\alpha \beta \gamma \] represents the constant in the equation.
So the equation \[{{x}^{3}}-7x+7=0\] is given in the question and we have to find the value of \[\left( \dfrac{1}{{{\alpha }^{4}}} \right)+\left( \dfrac{1}{{{\beta }^{4}}} \right)+\left( \dfrac{1}{{{\gamma }^{4}}} \right)\]
The coefficient of \[{{x}^{2}}\] is not given in the above equation so the value of
\[\alpha +\beta +\gamma =0\]
The value of \[\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{-7}{1}\]
\[\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =-7\]
The value of
\[\alpha \beta \gamma =-7\]
Now we have to find the value of the equation \[\left( \dfrac{1}{{{\alpha }^{4}}} \right)+\left( \dfrac{1}{{{\beta }^{4}}} \right)+\left( \dfrac{1}{{{\gamma }^{4}}} \right)\], using the above-given values which are as follows. On taking the LCM in the denominator, the equation becomes
\[\left( \dfrac{1}{{{\alpha }^{4}}} \right)+\left( \dfrac{1}{{{\beta }^{4}}} \right)+\left( \dfrac{1}{{{\gamma }^{4}}} \right)=\dfrac{{{\beta }^{4}}{{\gamma }^{4}}+{{\alpha }^{4}}{{\gamma }^{4}}+{{\alpha }^{4}}{{\beta }^{4}}}{{{\alpha }^{4}}{{\beta }^{4}}{{\gamma }^{4}}}\]………..eq(1)
We know that, the value of
\[{{\alpha }^{4}}{{\beta }^{4}}+{{\beta }^{4}}{{\gamma }^{4}}+{{\alpha }^{4}}{{\gamma }^{4}}={{({{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\alpha }^{2}}{{\gamma }^{2}})}^{2}}-2({{\alpha }^{2}}{{\beta }^{4}}{{\gamma }^{2}}+{{\beta }^{2}}{{\alpha }^{4}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{4}}{{\beta }^{2}})\]
On putting this value in eq(1), we get
\[\left( \dfrac{1}{{{\alpha }^{4}}} \right)+\left( \dfrac{1}{{{\beta }^{4}}} \right)+\left( \dfrac{1}{{{\gamma }^{4}}} \right)=\dfrac{{{({{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\alpha }^{2}}{{\gamma }^{2}})}^{2}}-2({{\alpha }^{2}}{{\beta }^{4}}{{\gamma }^{2}}+{{\beta }^{2}}{{\alpha }^{4}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{4}}{{\beta }^{2}})}{{{\alpha }^{4}}{{\beta }^{4}}{{\gamma }^{4}}}\]……..eq(2)
We also know that,
\[{{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\alpha }^{2}}{{\gamma }^{2}}={{(\alpha \beta +\beta \gamma +\gamma \alpha )}^{2}}-2(\alpha {{\beta }^{2}}\gamma +\beta {{\gamma }^{2}}\alpha +\gamma {{\alpha }^{2}}\beta )\]
On putting this value in eq(2), we get the following results
\[\left( \dfrac{1}{{{\alpha }^{4}}} \right)+\left( \dfrac{1}{{{\beta }^{4}}} \right)+\left( \dfrac{1}{{{\gamma }^{4}}} \right)=\dfrac{{{[{{(\alpha \beta +\beta \gamma +\gamma \alpha )}^{2}}-2(\alpha {{\beta }^{2}}\gamma +\beta {{\gamma }^{2}}\alpha +\gamma {{\alpha }^{2}}\beta )]}^{2}}-2{{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}({{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}})}{{{\alpha }^{4}}{{\beta }^{4}}{{\gamma }^{4}}}\]
…….eq(3)
Also, we know that,
\[{{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}={{(\alpha +\beta +\gamma )}^{2}}-2(\alpha \beta +\beta \gamma +\gamma \alpha )\]
On putting this value in eq(3), following results will be obtained
\[\left( \dfrac{1}{{{\alpha }^{4}}} \right)+\left( \dfrac{1}{{{\beta }^{4}}} \right)+\left( \dfrac{1}{{{\gamma }^{4}}} \right)=\dfrac{{{[{{(\alpha \beta +\beta \gamma +\gamma \alpha )}^{2}}-2\alpha \beta \gamma (\beta +\gamma +\alpha )]}^{2}}-2{{(\alpha \beta \gamma )}^{2}}({{(\alpha +\beta +\gamma )}^{2}}-2(\alpha \beta +\beta \gamma +\gamma \alpha ))}{{{\alpha }^{4}}{{\beta }^{4}}{{\gamma }^{4}}}\]……eq(4)
We know that, if the equation is given in the form of \[a{{x}^{3}}+b{{x}^{2}}+cx+d\] then their roots can be expressed in the form as given below
\[\begin{align}
  & \alpha +\beta +\gamma =-\dfrac{b}{a} \\
 & \alpha \beta +\beta \gamma +\alpha \gamma =\dfrac{c}{a} \\
 & \alpha \beta \gamma =-\dfrac{d}{a} \\
\end{align}\]
From the question, we know the following information that
\[\alpha +\beta +\gamma =0\]
\[\alpha \beta +\beta \gamma +\gamma \alpha =-7\]
\[\alpha \beta \gamma =-7\]
On putting these values in eq(4), the following result is obtained
\[\Rightarrow \left( \dfrac{1}{{{\alpha }^{4}}} \right)+\left( \dfrac{1}{{{\beta }^{4}}} \right)+\left( \dfrac{1}{{{\gamma }^{4}}} \right)=\dfrac{{{[{{(7)}^{2}}-0]}^{2}}-2{{(-7)}^{2}}(0-2(-7))}{{{(7)}^{4}}}\]
\[\Rightarrow \left( \dfrac{1}{{{\alpha }^{4}}} \right)+\left( \dfrac{1}{{{\beta }^{4}}} \right)+\left( \dfrac{1}{{{\gamma }^{4}}} \right)=\dfrac{{{(49)}^{2}}-2{{(-7)}^{2}}(0-2(-7))}{{{(49)}^{2}}}\]
\[\Rightarrow \left( \dfrac{1}{{{\alpha }^{4}}} \right)+\left( \dfrac{1}{{{\beta }^{4}}} \right)+\left( \dfrac{1}{{{\gamma }^{4}}} \right)=\dfrac{49[49-28]}{{{(49)}^{2}}}\]
\[\Rightarrow \left( \dfrac{1}{{{\alpha }^{4}}} \right)+\left( \dfrac{1}{{{\beta }^{4}}} \right)+\left( \dfrac{1}{{{\gamma }^{4}}} \right)=\dfrac{21}{49}\]
\[\Rightarrow \left( \dfrac{1}{{{\alpha }^{4}}} \right)+\left( \dfrac{1}{{{\beta }^{4}}} \right)+\left( \dfrac{1}{{{\gamma }^{4}}} \right)=\dfrac{3}{7}\]
So the above equation is the required answer.
So, the correct answer is “$\dfrac{3}{7}$”.

Note: If in a given polynomial, the highest degree of the variable is even then the arms of the graph of the polynomial will both be either upwards or both will be downwards and if in the given polynomial, the highest degree is odd then one arm of the graph will be up and one arm will be down.