Finding value of equation without solving for a quadratic equation How do I go about solving this problem: 
If $α$ and $β$ are the roots of $x^2+2x-3=0$, without solving the equation, find the values of $α^6 +β^6$.  
In my thoughts: I commenced by  expanding $(α +β)^6$, such that:
$$(α +β)^6 =α^6+6α^5β+15α^4β^2+20α^3β^3+15α^2β^4+6αβ^5+β^6$$ which when I reorganise:
$$(α +β)^6 =(α^6+β^6)+6α^5β+15α^4β^2+20α^3β^3+15α^2β^4+6αβ^5$$
when I isolate $(α^6+β^6)$ on one side:
$$(α^6+β^6) = (α +β)^6-6α^5β-15α^4β^2-20α^3β^3-15α^2β^4-6αβ^5$$
where does all this end for me to get a solution?
 A: HINT:
$$a^6+b^6=(a^3)^2+(b^3)^2=(a^3+b^3)^2-2(ab)^3\text{ and } a^3+b^3=(a+b)^3-3ab(a+b)$$
or
$$a^6+b^6=(a^2)^3+(b^2)^3=(a^2+b^2)^3-3(ab)^2(a^2+b^2) \text{ and } a^2+b^2=(a+b)^2-2ab$$
A: This exercise might be meant to make you realize that every symmetrical polynomial in $(\alpha,\beta)$ coincide with a (universal) polynomial in $(s,t)=(\alpha+\beta,\alpha\beta)$. For example, you might already be aware that
$$
\alpha^2+\beta^2=s^2-2t.
$$
Likewise,
$$
\alpha^6+\beta^6=s^6-6s^4t+9s^2t^2-2t^3.
$$
One can check that the polynomial on the RHS is homogeneous of degree $6$ provided one considers that the degree of $s$ is $1$ and the degree of $t$ is $2$.
In the case at hand, $s=-2$ and $t=-3$ hence
$$
\alpha^6+\beta^6=2^6+6\cdot2^4\cdot3+9\cdot2^2\cdot3^2+2\cdot3^3=730.
$$
More generally, one can obtain the expansion of $p_n=\alpha^n+\beta^n$ for every integer $n\geqslant0$ recursively, starting from $p_0=2$ and $p_1=s$, and using the relation
$$
p_{n+2}=sp_{n+1}-tp_n.
$$
Finally, note that, when $\alpha\beta\ne0$, one can also obtain the value of $p_n$ for negative values of $n$, using the identity
$$
p_{-n}=t^{-n}p_n.
$$
A: $\newcommand{\+}{^{\dagger}}%
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$\alpha\beta = -3\,,\quad \alpha + \beta = -2.\quad$ Let $\alpha > 0$ and
$\beta < 0.\quad$ Then, $\alpha\verts{\beta} = 3$ and
$\verts{\beta} - \alpha= 2.\quad$ Let $\alpha = 2\sinh^{2}\pars{\theta}.\quad$
$\verts{\beta} = 2\cosh^{2}\pars{\theta}$:
$$
3 = \bracks{2\sinh^{2}\pars{\theta}}\bracks{2\cosh^{2}\pars{\theta}} = \sinh^{2}\pars{2\theta}\quad\imp\quad\theta = \half{\rm arcsinh}\pars{\root{3}}
$$
