WAEC 2014 · Paper 2 · Q3

(a)
If $\alpha$ and $\beta$ are the roots of $3x^2 + 5x + 1 = 0$ , evaluate $27(\alpha^3 + \beta^3)$ .

Worked solution (try it first)

For $3x^2 + 5x + 1 = 0$ : $\alpha + \beta = -\frac{b}{a} = -\frac53$ and $\alpha\beta = \frac{c}{a} = \frac13$ .
Use $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$ .
Substitute: $\alpha^3 + \beta^3 = \left(-\frac53\right)^3 - 3\left(\frac13\right)\left(-\frac53\right)$ .
Work it out: $-\frac{125}{27} + \frac53 = -\frac{125}{27} + \frac{45}{27}$
$= -\frac{80}{27}$ .
Multiply by 27: $27(\alpha^3 + \beta^3) = -80$ .