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.