If \(\alpha\) and \(\beta\) are the roots of \(3x^{2} + 5x + 1 = 0\), evaluate \(27(\alpha^{3} + \beta^{3})\).
Question 1 Report
If \(\alpha\) and \(\beta\) are the roots of \(3x^{2} + 5x + 1 = 0\), evaluate \(27(\alpha^{3} + \beta^{3})\).
Answer Details
For \(3x^{2}+5x+1=0\) with roots \(\alpha,\beta\), the sum and product of roots are
\[\alpha+\beta=-\frac{5}{3},\qquad \alpha\beta=\frac{1}{3}.\]
Use the identity for the sum of cubes:
\[\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta).\]
Substitute the values:
\[\alpha^{3}+\beta^{3}=\left(-\frac{5}{3}\right)^{3}-3\left(\frac{1}{3}\right)\left(-\frac{5}{3}\right)=-\frac{125}{27}+\frac{5}{3}.\]
Writing over a common denominator of \(27\):
\[\alpha^{3}+\beta^{3}=-\frac{125}{27}+\frac{45}{27}=-\frac{80}{27}.\]
Therefore
\[27(\alpha^{3}+\beta^{3})=27\times\left(-\frac{80}{27}\right)=-80.\]
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