<s>[INST] 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?
 [/INST] 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
$$a^6+b^6=(a^2)^3+(b^2)^3=(a^2+b^2)^3-3(ab)^2(a^2+b^2) \text{ and } a^2+b^2=(a+b)^2-2ab$$
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