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(a) 2q

(b) ${{q}^{2}}+pr$

(c) ${{p}^{2}}-qr$

(d) r(pq-r)

Answer
Verified

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,

Complete step-by-step answer:

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 â€“

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

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.

Complete step-by-step answer:

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 â€“

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

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.

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