Question

If $a,b,c$ are the roots of the equation ${{x}^{3}}-p{{x}^{2}}+qx-r=0$, find the value of $\sum{\left( \dfrac{b}{c}+\dfrac{c}{b} \right)}$.

Answer 1

Hint: A cubic eq of the form $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$, where $a,b,c,d$ are non-zero, has roots as $\alpha ,\beta ,\gamma $ . We can solve this question using the relation between the three roots, which can be written as

$\alpha +\beta +\gamma =\dfrac{-b}{a}$

$\alpha \beta +\beta \lambda +\gamma \alpha =\dfrac{c}{a}$

$\alpha \beta \gamma =\dfrac{-d}{a}$

Complete step-by-step answer:
It is given in the question that $a,b,c$ are the roots of the equation ${{x}^{3}}-p{{x}^{2}}+qx-r=0$. We are given a cubic equation ${{x}^{3}}-p{{x}^{2}}+qx-r=0$ in the question. A cubic equation is the equation whose degree is 3 (degree is the highest power of a variable in the equation).

So we can write the relation between the roots of equation and coefficient of variable of equation. Let us consider that we have a cubic equation in the form $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$, where $a,b,c,d$ are non-zero. Then, we know that the cubic equation has three roots, so let $\alpha ,\beta ,\gamma $ be its roots. The relation between root and coefficient is given by

$\alpha +\beta +\gamma =\dfrac{-b}{a}$

$\alpha \beta +\beta \lambda +\gamma \alpha =\dfrac{c}{a}$

$\alpha \beta \gamma =\dfrac{-d}{a}$

Now, we have the cubic equation as ${{x}^{3}}-p{{x}^{2}}+qx-r=0$ and it is given that the roots of this equation are $a,b,c$. So, we can use the known relationships in the given equation and we will get

$a+b+c=p\ldots \ldots \ldots \left( i \right)$

$ab+bc+ca=q\ldots \ldots \ldots \left( ii \right)$

$abc=r\ldots \ldots \ldots \left( iii \right)$

Now, we need to find $\sum{\left( \dfrac{b}{c}+\dfrac{c}{b} \right)}$. It means that we need to find the summation of $\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{c}{a}+\dfrac{a}{c}$. We need to change the form of the equation to get the answer. So we will divide equation (ii) by equation (iii), and we will get

$\dfrac{ab+bc+ca}{abc}=\dfrac{q}{r}$

By separating the denominator in left-hand side of the equation, we will get

$\dfrac{ab}{abc}+\dfrac{bc}{abc}+\dfrac{ac}{abc}=\dfrac{q}{r}$

We can observe that $ab,bc,ac$ are cancelling out from each term respectively in the left-hand side of the equation. Then, we will get,

$\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{q}{r}\ldots \ldots \ldots \left( iv \right)$

Now we will multiply equation (iv) and equation (i) and we will get,

$\left( a+b+c \right)\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \right)=\dfrac{pq}{r}$

Now we will open the bracket and simplify it further as below,

$1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1=\dfrac{pq}{r}$

Now we will rearrange the terms to get the desired answer as,

$\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{a}{c}+\dfrac{c}{a}+3=\dfrac{pq}{r}$

The left-hand side of the equation can be written as $\sum{\left( \dfrac{b}{c}+\dfrac{c}{b} \right)}$. So, the equation becomes, $\sum{\left( \dfrac{b}{c}+\dfrac{c}{b} \right)}+3=\dfrac{pq}{r}$

Rearranging, we will get it as,

$\begin{align}

  & \sum{\left( \dfrac{b}{c}+\dfrac{c}{b} \right)}=\dfrac{pq}{r}-3 \\

 & \sum{\left( \dfrac{b}{c}+\dfrac{c}{b} \right)}=\dfrac{pq-3r}{r} \\

\end{align}$

Hence, the final answer is $\sum{\left( \dfrac{b}{c}+\dfrac{c}{b} \right)}=\dfrac{pq-3r}{r}$.


Note: The possibility of mistake that could be done in the question is while finding the value of the expression given in question i.e. $\sum{\left( \dfrac{b}{c}+\dfrac{c}{b} \right)}$ . It could be wrongly interpreted as to find only the sum of $\dfrac{b}{c}+\dfrac{c}{b}$. But we have to find the sum of $\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{c}{a}+\dfrac{a}{c}$.