# Find $\alpha^3 + \beta^3$ which are roots of a quadratic equation.

I have a question.

Given a quadratic polynomial, $ax^2 +bx+c$, and having roots $\alpha$ and $\beta$. Find $\alpha^3+\beta^3$. Also find $\frac1\alpha^3+\frac1\beta^3$

I don't know how to proceed. Any help would be appreciated.

• Don't you mean $ax^2+bx+c$? – vrugtehagel Jan 29 '16 at 9:58
• Step 1: find $\alpha+\beta$ and $\alpha\beta$. Step 2: Find an expression for $\alpha^3+\beta^3$ in terms of $\alpha+\beta$ and $\alpha\beta$. For that, you'll want to look at things like $(\alpha+\beta)^3$ and $\alpha\beta(\alpha+\beta)$ and suchlike. – Gerry Myerson Jan 29 '16 at 10:03

First note that $\alpha^3+\beta^3=(\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2)$ and also note that $-\frac{b}{a}=\alpha+\beta$ and $\frac{c}{a}=\alpha\beta$ (do you see why?) We can make $$\alpha^2++2\alpha\beta+\beta^2=(\alpha+\beta)^2=\frac{b^2}{a^2}$$ so our final outcome will be \begin{align} \alpha^3+\beta^3&=(\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2)\\ &= -\frac{b}{a}(\alpha^2+2\alpha\beta+\beta^2-3\alpha\beta)\\ &= -\frac{b}{a}(\frac{b^2}{a^2}-3\frac{c}{a})\\ &= -\frac{b^3-3abc}{a^3} \end{align}

Hope this helped!

• I see that $−b/a=α+β=α+β$ and $ca=αβ$ can be derived by manipulating the quadratic formula. – TheRandomGuy Jan 29 '16 at 10:05
• You can also see $a(x-\alpha)(x-\beta)=ax^2-a\cdot(\alpha+\beta)x+a\cdot\alpha\beta=ax^2+bx+c$ so that $b=-a\cdot(\alpha+\beta)$ and $c=a\cdot\alpha\beta$ – vrugtehagel Jan 29 '16 at 10:08
• Great! but that's the factored form. Oh! So those formulas can be derived using it. – TheRandomGuy Jan 29 '16 at 10:11

Use Viete formulas:

$$\alpha\beta = c/a$$$$\alpha + \beta = - b/a$$

Therefore $$\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha^2\beta - 3\alpha\beta^2 = (-b/a)^3 + 3bc/a^2$$

For $\frac1{\alpha^3}+\frac1{\beta^3}$, use that the roots of $a+bx+cx^2$ are $\frac1\alpha$ and $\frac1\beta$ to reduce to the previous problem.

Just to be different.

If $\alpha$ is a solution of $ax^2 + bx + c = 0$ Then

$a\alpha^2 + b\alpha + c = 0$

So
$\quad \alpha^2 = -\dfrac{b\alpha + c}{a}$
$\quad \alpha^3 = -\dfrac{b\alpha^2 + c\alpha}{a} = -\dfrac{b\left( -\dfrac{b\alpha + c}{a} \right)+ c\alpha}{a} = -\dfrac{ -b^2\alpha - bc + ac\alpha}{a^2} = -\dfrac{(ac -b^2)\alpha - bc}{a^2}$

Similarly,
$\quad \beta^3 = -\dfrac{(ac -b^2)\beta - bc}{a^2}$

So $\alpha^3 + \beta^3 = -\dfrac{(ac -b^2)(\alpha + \beta) - 2bc}{a^2} = \dfrac{(b^2 - ac)(\alpha + \beta) + 2bc}{a^2}$

Since $\alpha + \beta = -\dfrac ba$, we see that

$\alpha^3 + \beta^3 = \dfrac{(b^2 - ac) \left(-\dfrac ba \right) + 2bc}{a^2} = \dfrac{(-b^3 + abc) + 2abc}{a^3} = \dfrac{3abc - b^3}{a^3}$