Given $\alpha$ and $\beta$ are the roots of the quadratic equation $6x^2 + 2x – 3 = 0$, how do I find the value of:
$$
\alpha^3 + \beta^3
$$
and:
$$
\frac{1}{\alpha^3} + \frac{1}{\beta^3}
$$
Taking the product and sum rule, I know that:
$$
\alpha + \beta = -\frac{1}{3}
$$
$$
\alpha.\beta = -\frac{1}{2}
$$
And I tried doing:
$$
\alpha^3 + \beta^3 = (\alpha + \beta)^3 – 3\alpha^2\beta – 3\alpha\beta^2
$$
$$
= -\frac{1}{9}-3\alpha(-\frac{1}{2})-3\beta(-\frac{1}{2})
$$
$$
= -\frac{1}{9}+\left(\frac{3\alpha}{2}\right)+\left(\frac{3\beta}{2}\right)
$$
But I know this is wrong because the answer is apparently $-\frac{29}{54}$. Can anyone help?
Best Answer
Hint: $(\alpha + \beta)^3 = \alpha^3 + \beta^3 + 3\alpha \beta (\alpha + \beta)$
You're basically there with $\displaystyle\alpha^3 + \beta^3 = -\frac1{27} + \frac{3\alpha}2 + \frac{3\beta}2 = -\frac1{27} + \frac32(\alpha + \beta)=\ldots$
(Note that it's $\frac{1}{27}$ not $\frac19$)
And $\displaystyle \frac1{\alpha^3} + \frac1{\beta^3} = \frac{\alpha^3 + \beta^3}{(\alpha \beta)^3}$