If α and β are the roots of the equation $3x^2+5x+4=0$, find the value of $α^3+β^3$

Question

If α and β are the roots of the equation $3x^2+5x+4=0$, find the value of $α^3+β^3$

How can I factorize the expression to use the rule of sum and product of roots?

The answer is $\frac{55}{27}$

Answer 1

$\newcommand{\al}{\alpha}\newcommand{\be}\beta$ I hope you can find $u=\al+\be$ and $v=\al\be$. It would be nice if $u^3$ was equal to $\al^3+\be^3$, but it isn't. Actually $$u^3=\al^3+\be^3+3\al^2\be+3\al\be^2.$$ We have a deficit of $$3\al^2\be+3\al\be^2=3\al\be(\al+\be)=3uv.$$ Therefore $$u^3=\al^3+\be^3+3uv,$$ or $$\al^3+\be^3=u^3-3uv.$$

Answer 2

Vieta's theorem is the key, like in your previous question. It gives $\alpha+\beta=-\frac{5}{3}$ with no effort and

$$\alpha^3+\beta^3 = (\alpha+\beta)\left((\alpha+\beta)^2-3\alpha\beta\right) = -\frac{5}{3}\left(\frac{25}{9}-4\right) = \color{red}{\frac{55}{27}}$$ with very simple manipulations.

Answer 3

Using the general form of quadritic equation, $x^2 -(a+b) x + ab$, we get the values of $a+b$ and $ab$.

Now, the expression $a^3+b^3$ can be reduced to $(a+b)^3 -3ab(a+b)$.

Substitute the value of $a+b$ and $ab$ in the above equation.

Answer 4

Rename $a=\alpha$ and $b=\beta$

Since $a$ is a solution of $3x^2+5x+4=0$ we have $3a^2+5a+4=0$ and thus $a^2={1\over 3}(-5a-4)$, so:

$$ 3a^3 = -5a^2-4a = -{5\over 3}(-5a-4) -4a = {13\over 3}a+{20\over 3}$$

the same is true for $b$, so we have

$$9a^3+9b^3 = 13(a+b)+40 = 13{-5\over 3 }+40 = {55\over 3}$$

and thus conclusion.

Answer 5

expanding $$\frac{1}{216} \left(-5-i \sqrt{23}\right)^3+\frac{1}{216} \left(-5+i \sqrt{23}\right)^3$$ we obtain $$\frac{55}{27}$$