Find the roots of $3x^3-4x-8$

Question

It is given that $\alpha$, $\beta$ and $\gamma$ are the roots of the polynomial $3x^3-4x-8$.

I have been asked to calculate the value of $\alpha^2 + \beta^2 + \gamma^2$.

However I am unsure how to find these roots, seeing as though I haven't been given a root to start with.

I began by identifying that

$\alpha + \beta + \gamma = 0$

$\alpha\beta + \alpha\gamma + \beta\gamma = -4/3$

$\alpha\beta\gamma = 8/3$

However I am unsure how to continue to find $\alpha^2 + \beta^2 + \gamma^2$.

Answer 1

Use the identity $(a+b+c)^{2}=a^{2} + b^{2} + c^{2} + 2ab+ 2bc+2ca$

Answer 2

$ x(3x^2-4)=8$

Squaring we get $x^2(3x^2-4)^2=8^2$

Let $x^2=y\implies y(3y-4)^2=64\iff9y^3-24y^2+16y-64=0$

whose roots are $a^2,b^2,c^2$

$\implies a^2+b^2+c^2=\dfrac{24}9=?$

Answer 3

Alternative approach: for any root we have $3x^3-4x=8$, hence by squaring $9x^6-24x^4+16 x^2 = 64$ and $\alpha^2,\beta^2,\gamma^2$ are the roots of the polynomial $9z^3-24z^2+16z-64$.

By Vièta's formulas it follows that $\alpha^2+\beta^2+\gamma^2 = \frac{24}{9} = \color{red}{\frac{8}{3}}$.

Answer 4

Yet another approach: $\,\dfrac{1}{\alpha}, \dfrac{1}{\beta}, \dfrac{1}{\gamma}\,$ are the roots of $\,8x^3+4x^2-3\,$, so $\,\dfrac{1}{\alpha}+ \dfrac{1}{\beta}+ \dfrac{1}{\gamma}=-\dfrac{1}{2}\,$.

But $\,3\alpha^3=4\alpha+8 \iff \alpha^2=\dfrac{4}{3} + \dfrac{8}{3\alpha}\,$, and therefore: $$\alpha^2+\beta^2+\gamma^2=3 \cdot \frac{4}{3}+ \frac{8}{3}\left(\frac{1}{\alpha}+ \frac{1}{\beta}+ \frac{1}{\gamma}\right)=4-\frac{4}{3} = \frac{8}{3}$$