Show that $\alpha^3 = 10\alpha - 24$ given that $\alpha + \beta = 4$ and $\alpha \beta = 6$

Question

How to show that $\alpha^3 = 10\alpha - 24$ given that $\alpha + \beta = 4$ and $\alpha \beta = 6$

As given in the title. I tried shifting the RHS to the left but I'm not sure if factorization is the way to go. There are earlier parts to this question but I'm not sure if they will be helpful so I'm excluding them.

Thank you.

(Question is under sum and product of roots for a quadratic equation.)

Answer 1

Note that $\alpha$ and $\beta$ are roots of the quadratic equation: $z^2-(\text{ Sum of Roots }) + \text{ Product of Roots } = 0 \implies z^2-4z +6 =0$.

Note as, $\alpha^2-4\alpha + 6=0$ (because $\alpha$ satisfies the quadratic), we can easily see that : $$\alpha^3 = 4\alpha^2 - 6\alpha = 4(4\alpha -6)-6\alpha = 10\alpha -24$$

Answer 2

Hint:

So, $\alpha$ is a root of $$t^2-4t+6=0$$

Now $$t^3-10t+24=(t^2-4t+6)(t-?)=0$$

Answer 3

Because $$\alpha+\frac{6}{\alpha}=4$$ or $$\alpha^2=4\alpha-6,$$ which gives $$\alpha^3=4\alpha^2-6\alpha=4(4\alpha-6)-6\alpha=10\alpha-24.$$

Answer 4

Hint:  given their sum and product, $\;\alpha,\beta$ are the roots of $z^2 - 4z + 6 = 0\,$.

It follows that $\alpha^2=4 \alpha-6\,$, then $\alpha^3 = 4 \alpha^2 - 6 \alpha = 4(4\alpha-6) - 6 \alpha= \cdots$

Answer 5

Sum of roots = $ \alpha + \beta + \gamma =$

$ \frac{-b}{a} = \frac{-0}{1} = 0 $

But, $ \alpha + \beta = 4 $ $\implies \gamma = -4 $

Product of roots $\alpha * \beta * \gamma = $

$ \frac{-d}{a} = -24 $

Sum of roots taken two at a time = $ \alpha\beta + \beta\alpha + \gamma\alpha =$

$ \frac{c}{a} = \frac{10}{1} = 10 $

Now, for a cubic equation: $$ x^3 - (\alpha+\beta+\gamma)x^2+(\alpha\beta + \beta\alpha + \gamma\alpha )x - (\alpha\beta\gamma)=0 $$ Or $$ \alpha^3-10\alpha+24=0 $$