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# Find the cubic equation whose roots are the cubes of the roots of x^3+ax^2+bx+c=0,a,b,cinRR?

## Find the cubic equation whose roots are the cubes of the roots of ${x}^{3} + a {x}^{2} + b x + c = 0$,$a , b , c \in \mathbb{R}$

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Mar 25, 2017

${x}^{3} + \left({a}^{3} - 3 a b + 3 c\right) {x}^{2} + \left({b}^{3} - 3 a b c + 3 {c}^{2}\right) x + {c}^{3} = 0$

#### Explanation:

Suppose the roots of the original cubic are $\alpha$, $\beta$ and $\gamma$.

Then:

${x}^{3} + a {x}^{2} + b x + c = \left(x - \alpha\right) \left(x - \beta\right) \left(x - \gamma\right)$

$\textcolor{w h i t e}{{x}^{3} + a {x}^{2} + b x + c} = {x}^{3} - \left(\alpha + \beta + \gamma\right) {x}^{2} + \left(\alpha \beta + \beta \gamma + \gamma \alpha\right) x - \alpha \beta \gamma$

So we have:

$\left\{\begin{matrix}\alpha + \beta + \gamma = - a \\ \alpha \beta + \beta \gamma + \gamma \alpha = b \\ \alpha \beta \gamma = - c\end{matrix}\right.$

The cubic we are looking for is:

$\left(x - {\alpha}^{3}\right) \left(x - {\beta}^{3}\right) \left(x - {\gamma}^{3}\right)$

$= {x}^{3} - \left({\alpha}^{3} + {\beta}^{3} + {\gamma}^{3}\right) {x}^{2} + \left({\alpha}^{3} {\beta}^{3} + {\beta}^{3} {\gamma}^{3} + {\gamma}^{3} {\alpha}^{3}\right) x - {\alpha}^{3} {\beta}^{3} {\gamma}^{3}$

So the problem essentially boils down to expressing each of the symmetric polynomials in ${\alpha}^{3}$, ${\beta}^{3}$ and ${\gamma}^{3}$ constituting these coefficients in terms of the elementary symmetric polynomials in $\alpha$, $\beta$ and $\gamma$.

For example:

${\left(\alpha + \beta + \gamma\right)}^{3}$

$= {\alpha}^{3} + {\beta}^{3} + {\gamma}^{3} + 3 \left({\alpha}^{2} \beta + {\beta}^{2} \gamma + {\gamma}^{2} \alpha + \alpha {\beta}^{2} + \beta {\gamma}^{2} + \gamma {\alpha}^{2}\right) + 6 \alpha \beta \gamma$

$\left(\alpha + \beta + \gamma\right) \left(\alpha \beta + \beta \gamma + \gamma \alpha\right)$

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

So:

${\alpha}^{3} + {\beta}^{3} + {\gamma}^{3}$

$= {\left(\alpha + \beta + \gamma\right)}^{3} - 3 \left(\alpha + \beta + \gamma\right) \left(\alpha \beta + \beta \gamma + \gamma \alpha\right) + 3 \alpha \beta \gamma$

$= - {a}^{3} + 3 a b - 3 c$

We also find:

${\left(\alpha \beta + \beta \gamma + \gamma \alpha\right)}^{3}$

$= {\alpha}^{3} {\beta}^{3} + {\beta}^{3} {\gamma}^{3} + {\gamma}^{3} {\alpha}^{3} + 3 \left({\alpha}^{3} {\beta}^{2} \gamma + {\beta}^{3} {\gamma}^{2} \alpha + {\gamma}^{3} {\alpha}^{2} \beta + {\alpha}^{3} \beta {\gamma}^{2} + {\beta}^{3} \gamma {\alpha}^{2} + {\gamma}^{3} \alpha {\beta}^{2}\right) + 6 {\alpha}^{2} {\beta}^{2} {\gamma}^{2}$

$\left(\alpha + \beta + \gamma\right) \left(\alpha \beta + \beta \gamma + \gamma \alpha\right) \alpha \beta \gamma$

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

So:

${\alpha}^{3} {\beta}^{3} + {\beta}^{3} {\gamma}^{3} + {\gamma}^{3} {\alpha}^{3}$

$= {\left(\alpha \beta + \beta \gamma + \gamma \alpha\right)}^{3} - 3 \left(\alpha + \beta + \gamma\right) \left(\alpha \beta + \beta \gamma + \gamma \alpha\right) \alpha \beta \gamma + 3 {\left(\alpha \beta \gamma\right)}^{2}$

$= {b}^{3} - 3 a b c + 3 {c}^{2}$

Finally:

${\alpha}^{3} {\beta}^{3} {\gamma}^{3} = {\left(\alpha \beta \gamma\right)}^{3} = - {c}^{3}$

Hence the required cubic equation is:

${x}^{3} + \left({a}^{3} - 3 a b + 3 c\right) {x}^{2} + \left({b}^{3} - 3 a b c + 3 {c}^{2}\right) x + {c}^{3} = 0$

$\textcolor{w h i t e}{}$
Footnote

If you would like to see a more advanced application of symmetric polynomials, you may like to take a look at this one: https://socratic.org/s/aCWXbG2b

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