WAEC 2022 · Paper 2 · Q2

If $\alpha$ and $\beta$ are the roots of $2x^2 - x - 2 = 0$ , find the value of $\alpha^3 + \beta^3$ .

Worked solution (try it first)

For $2x^2 - x - 2 = 0$ : $\alpha + \beta = -\frac{-1}{2} = \frac12$ and $\alpha\beta = \frac{-2}{2} = -1$ .
Write the cube in terms of them: $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$ .
First part: $\left(\frac12\right)^3 = \frac18$ .
Second part: $3\alpha\beta(\alpha + \beta) = 3(-1)\left(\frac12\right)$
$= -\frac32$ .
So $\alpha^3 + \beta^3 = \frac18 - \left(-\frac32\right)$
$= \frac18 + \frac{12}{8}$
$= \frac{13}{8}$ .