If $\alpha$ and $\beta$ are the roots of the quadratic equation $2x^2 + 4x -5 = 0$, evaluate $\alpha^3 + \beta^3$..

I know that $$\alpha + \beta = \frac{-b}{a}$$ and $$\alpha \beta = \frac{c}{a}\\$$ I also know that factorising $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha + \beta -\alpha \beta)$

I substituted everything in..

$$ (\frac{-4}{2})((\frac{-4}{2})-(\frac{-5}{2})) = -1 $$

Unfortunately, the answer in my text book says otherwise, it's supposed to be $-23$. What am I doing wrong? Thanks in advance!

  • 3
    $\begingroup$ Your factorisation is wrong, $\alpha^3+\beta^3 = (\alpha+\beta)(\alpha^2+\beta^2-\alpha\beta)$. But easier: $\alpha^3+\beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$. $\endgroup$ – Daniel Fischer Oct 20 '14 at 12:21
  • $\begingroup$ Awkward.. Thank you! Well that was a stupid mistake.. Thanks again. $\endgroup$ – Samir Chahine Oct 20 '14 at 12:22

You have that $$\begin{align*}α^3+β^3&=(α+β)(α^2-αβ+β^2)=(α+β)(α^2+2αβ+β^2-3αβ)=\\&=(α+β)((α+β)^2-3αβ)\end{align*}$$


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