Question

If the sum of the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is equal to the sum of the cubes of their reciprocals, then prove that $a{{b}^{2}}=3{{a}^{2}}c+{{c}^{3}}$.

Answer 1

Hint: In this question, we are given the form of the quadratic equation and the sum of the cubes of the reciprocals of the roots. Therefore, we can use the standard formulas for finding the sum and product of the roots of a quadratic equation and then use the given relation to derive the expression given in the question.

Complete step-by-step solution -
We are given that the quadratic equation is $a{{x}^{2}}+bx+c=0$. Therefore, from the theory of quadratic equations, if $\alpha $ and $\beta $ are the roots of this equation, then their sum and product should be given by
$\alpha +\beta =\dfrac{-b}{a}......................(1.1)$
and
$\alpha \beta =\dfrac{c}{a}.......................(1.2)$
Now, it is also given that the sum of the roots is equal to the sum of the cubes of their reciprocals, therefore, we obtain
$\begin{align}
  & \alpha +\beta =\dfrac{1}{{{\alpha }^{3}}}+\dfrac{1}{{{\beta }^{3}}} \\
 & \Rightarrow \alpha +\beta =\dfrac{{{\beta }^{3}}+{{\alpha }^{3}}}{{{\left( \alpha \beta \right)}^{3}}}......................(1.3) \\
\end{align}$
Now, we know that the formula for cube of the sum of two numbers is given by
$\begin{align}
  & {{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right) \\
 & \Rightarrow {{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right).............(1.4) \\
\end{align}$
Using this in equation (1.3), we obtain
$\alpha +\beta =\dfrac{{{\left( \alpha +\beta \right)}^{3}}-3\alpha \beta \left( \alpha +\beta \right)}{{{\left( \alpha \beta \right)}^{3}}}$
Now, we can use the values from (1.1) and (1.2) to get
$\dfrac{-b}{a}=\dfrac{{{\left( \dfrac{-b}{a} \right)}^{3}}-3\times \dfrac{c}{a}\times \dfrac{-b}{a}}{{{\left( \dfrac{c}{a} \right)}^{3}}}=\dfrac{-b}{a}\times \left( \dfrac{{{\left( \dfrac{-b}{a} \right)}^{2}}-3\times \dfrac{c}{a}}{{{\left( \dfrac{c}{a} \right)}^{3}}} \right)$
We can now cancel out $\dfrac{-b}{a}$ from both sides to obtain
$\begin{align}
  & \dfrac{{{\left( \dfrac{-b}{a} \right)}^{2}}-3\times \dfrac{c}{a}}{{{\left( \dfrac{c}{a} \right)}^{3}}}=1 \\
 & \Rightarrow {{\left( \dfrac{-b}{a} \right)}^{2}}-3\times \dfrac{c}{a}={{\left( \dfrac{c}{a} \right)}^{3}} \\
 & \Rightarrow \dfrac{{{b}^{2}}-3ca}{{{a}^{2}}}=\dfrac{{{c}^{3}}}{{{a}^{3}}} \\
\end{align}$
Now, we can cancel out ${{a}^{2}}$ form both sides of the denominator to obtain
\[\begin{align}
  & {{b}^{2}}-3ac=\dfrac{{{c}^{3}}}{a} \\
 & \Rightarrow a{{b}^{2}}-3{{a}^{2}}c={{c}^{3}} \\
\end{align}\]
Taking the second term in the LHS to the RHS, we obtain
$a{{b}^{2}}=3{{a}^{2}}c+{{c}^{3}}$
Which is what we wanted to prove.

Note: We should note that while canceling out terms on both sides, we should make sure that the terms being canceled are not equal to zero. However, in this general case, we can assume that $a$ and $\dfrac{-b}{a}$ are not equal to zero and thus the proof remains valid in this case.