If α and β are the roots of the equation $3x^2+5x+4=0$, find the value of $α^3+β^3$ 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 $\frac{55}{27}$
 A: $\newcommand{\al}{\alpha}\newcommand{\be}\beta$ 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
$$\al^3+\be^3=u^3-3uv.$$
A: 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.
A: 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.
A: 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.   
A: expanding $$\frac{1}{216} \left(-5-i
   \sqrt{23}\right)^3+\frac{1}{216} \left(-5+i
   \sqrt{23}\right)^3$$ we obtain $$\frac{55}{27}$$
