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# If $\alpha ,\text{ }\beta \text{ and }\gamma$ are the roots of the equation ${{x}^{3}}+p{{x}^{2}}+qx+r$=0, then the coefficient of x in the cubic equation whose roots are $\alpha (\beta +\gamma ),\beta (\gamma +\alpha )\text{ and }\gamma (\alpha +\beta )$ is?(a) 2q(b) ${{q}^{2}}+pr$(c) ${{p}^{2}}-qr$(d) r(pq-r) Verified
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Hint: To solve this problem, we make use of the basic properties of a cubic polynomial related to the relation of sum of roots, product of roots and product of roots taken two at a time. That is,

For, ${{x}^{3}}+p{{x}^{2}}+qx+r$=0,

$\alpha +\beta +\gamma$=-p

$\alpha \beta +\beta \gamma +\gamma \alpha$=q

$\alpha \beta \gamma$=-r

We make use of these properties to find the cubic equation with new roots.

We have the roots as $\alpha (\beta +\gamma ),\beta (\gamma +\alpha )\text{ and }\gamma (\alpha +\beta )$. We know how to find the sum of roots, product of roots and product of roots taken two at a time in case of these new roots.

In the question in particular, we need to find the coefficient of x for the cubic polynomial-
${{x}^{3}}+p{{x}^{2}}+qx+r$=0

Coefficient of x is given by product of roots of the cubic equation taken two at a time. In the normal cubic polynomial, this was $\alpha \beta +\beta \gamma +\gamma \alpha$.

In case of the new roots, $\alpha (\beta +\gamma ),\beta (\gamma +\alpha )\text{ and }\gamma (\alpha +\beta )$, the product of the new roots taken two at a time is –
=$\alpha \beta (\beta +\gamma )(\gamma +\alpha )\text{ + }\beta \gamma (\gamma +\alpha )(\alpha +\beta )+\gamma \alpha (\alpha +\beta )(\beta +\gamma )$
We now expand each of these three terms, we get,

=$[{{\alpha }^{2}}{{\beta }^{2}}\text{+}{{\alpha }^{2}}\beta \gamma +\alpha {{\beta }^{2}}\gamma +\alpha \beta {{\gamma }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\alpha }^{2}}\beta \gamma +\alpha {{\beta }^{2}}\gamma +\alpha \beta {{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}+{{\alpha }^{2}}\beta \gamma +\alpha {{\beta }^{2}}\gamma +\alpha \beta {{\gamma }^{2}}$]

=${{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}\text{+3}{{\alpha }^{2}}\beta \gamma +3\alpha {{\beta }^{2}}\gamma +3\alpha \beta {{\gamma }^{2}}$

=${{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}\text{+3}\alpha \beta \gamma (\alpha +\beta +\gamma )$-- (1)

Now, we know that,

q=$\alpha \beta +\beta \gamma +\gamma \alpha$

Squaring LHS and RHS, we get,

${{q}^{2}}=$${{\left( \alpha \beta +\beta \gamma +\gamma \alpha \right)}^{2}}$

We use the following algebraic identity,

${{(a+b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2(ab+bc+ca)$
Thus, we have,

${{q}^{2}}=$${{(\alpha \beta )}^{2}}+{{(\beta \gamma )}^{2}}+{{(\gamma \alpha )}^{2}}+2(\alpha \beta )(\beta \gamma )+2(\beta \gamma )(\gamma \alpha )+2(\gamma \alpha )(\alpha \beta )$

Thus, be re-arranging, we would have,

${{q}^{2}}=$${{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}\text{+2}\alpha \beta \gamma (\alpha +\beta +\gamma )$-- (2)

Now, from (1)

$\alpha \beta (\beta +\gamma )(\gamma +\alpha )\text{ + }\beta \gamma (\gamma +\alpha )(\alpha +\beta )+\gamma \alpha (\alpha +\beta )(\beta +\gamma )$=${{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}\text{+3}\alpha \beta \gamma (\alpha +\beta +\gamma )$

Thus,

${{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}\text{+3}\alpha \beta \gamma (\alpha +\beta +\gamma )$=${{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}\text{+2}\alpha \beta \gamma (\alpha +\beta +\gamma )+\alpha \beta \gamma (\alpha +\beta +\gamma )$
From (2), we have,

${{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}\text{+3}\alpha \beta \gamma (\alpha \beta +\beta \gamma +\gamma \alpha )$= ${{q}^{2}}+(-p)(-r)$

Since, we know that,

$\alpha +\beta +\gamma$=-p

$\alpha \beta \gamma$=-r

Thus, we have,

${{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}\text{+3}\alpha \beta \gamma (\alpha \beta +\beta \gamma +\gamma \alpha )$=${{q}^{2}}+pr$

Hence, the correct answer is (b) ${{q}^{2}}+pr$.

Note: For solving problems related to roots of a cubic polynomial, we should know the basic properties of sum of roots, product of roots and product of roots taken two at a time. Further, it is then to solve the problem, one should be aware about the basic manipulations involving algebraic terms like re-grouping, re-arrangement and usage of the known properties.

Last updated date: 21st Sep 2023
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