2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Don't you mean $ax^2+bx+c$? $\endgroup$ – vrugtehagel Jan 29 '16 at 9:58
  • $\begingroup$ 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. $\endgroup$ – Gerry Myerson Jan 29 '16 at 10:03
4
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ I see that $−b/a=α+β=α+β$ and $ca=αβ$ can be derived by manipulating the quadratic formula. $\endgroup$ – TheRandomGuy Jan 29 '16 at 10:05
  • $\begingroup$ 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 $c=a\cdot\alpha\beta$ $\endgroup$ – vrugtehagel Jan 29 '16 at 10:08
  • $\begingroup$ Great! but that's the factored form. Oh! So those formulas can be derived using it. $\endgroup$ – TheRandomGuy Jan 29 '16 at 10:11
6
$\begingroup$

Use Viete formulas:

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

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.