If \(\alpha\) and \(\beta\) are the roots of \(3x^{2} + 5x + 1 = 0\),

FURTHER MATHEMATICS
WAEC 2014

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

Explanation

\(3x^{2} + 5x + 1 = 0\)

\(x^{2} + \frac{5}{3}x + \frac{1}{3} = 0\)

\(\alpha \beta = \frac{c}{a}\)

= \(\frac{1}{3}\)

\(\alpha + \beta = \frac{-b}{a}\)

= \(-\frac{5}{3}\)

\((\alpha + \beta)^{3} = \alpha^{3} + 3\alpha^{2} \beta + 3\alpha \beta^{2} + \beta^{3}\)

\(\therefore \alpha^{3} + \beta^{3} = (\alpha + \beta)^{3} - 3\alpha^{2} \beta - 3\alpha \beta^{2}\)

= \((\alpha + \beta)^{3} - 3\alpha \beta (\alpha + \beta)\)

= \((-\frac{5}{3})^{3} - 3(\frac{1}{3})(-\frac{5}{3})\)

= \((-\frac{125}{27}) + \frac{5}{3}\)

\(\therefore (\alpha^{3} + \beta^{3}) = -\frac{80}{27}\)

\(27(\alpha^{3} + \beta^{3}) = 27(-\frac{80}{27})\)

= -80.



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