Using sum/product of quadratic roots to solve cubic equation

Question

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?

Answer 1

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 $\displaystyle \frac1{\alpha^3} + \frac1{\beta^3} = \frac{\alpha^3 + \beta^3}{(\alpha \beta)^3}$

Answer 2

Since $\alpha +\beta=-\frac{1}{3}$

$\alpha \cdot \beta=-\frac{1}{2}$

The value of $\alpha^3+\beta^3={(\alpha +\beta)}^3-3\alpha\beta \cdot (\alpha+\beta)$

$\implies (\frac{-1}{3})^3+3\cdot\frac{1}{2}\cdot (\frac{-1}{3})$

$\implies \frac{-29}{54}$

For the second question, the value would be $\frac{\frac{-29}{54}}{(\frac{-1}{2})^3}$ $\implies \frac{116}{27}$

Answer 3

We have $6x^2+2x=3$

Cubing both sides we get $\displaystyle(6x^2)^3+(2x)^3+3(6x^2)(2x)(6x^2+2x)=3^3$

$\displaystyle\iff216(x^3)^2+8(x^3)+36(x^3)(3)=27$ as $6x^2+2x=3$

Setting $\displaystyle x^3=y,216y^3+116y-27=0$ whose roots are $\alpha^3,\beta^3$

So, we have $\displaystyle \alpha^3+\beta^3=-\frac{116}{216}=\cdots$ and $\displaystyle \alpha^3\beta^3=-\frac{27}{216}=\cdots$

Finally, $\displaystyle \frac1{\alpha^3}+\frac1{\beta^3}=\frac{\alpha^3+\beta^3}{\alpha^3\beta^3}=\cdots$

Answer 4

I'm going to use a and b for easier explanation. (A + B)^3 does not equal A^3 + B^3. It is the addition of two cubes, which is equal to: (A+B)(A^2-AB+B^2). You must do that with your alpha and beta. You should have: Line 1: A^3 + B^3 Line 2: (A+B)(A^2-AB+B^2) Line 3: (A+B)(A^2-B^2-AB) And you should be able to work out all those separately and substitute each answer in.