# 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?

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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 $$p_{-n}=t^{-n}p_n.$$

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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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sorry I had left out the coefficients in my original question. Now added –  Sylvester Nov 27 '13 at 9:22
@Sylvester, though that method is not the best, we can use $$6α^5β+15α^4β^2+20α^3β^3+15α^2β^4+6αβ^5=6\alpha\beta(\alpha^4+\beta^4)+20(\alph‌​a \beta)^3+15(\alpha\beta)^2(\alpha^2+\beta^2)$$ Then $\alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2(\alpha\beta)^2$ –  lab bhattacharjee Nov 27 '13 at 9:56


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wouldnt $\alpha+\beta=-2$? –  wfw00d Nov 27 '13 at 10:25
@wfw00d Yes. You are right. I already rewrote the answer. Thanks. –  Felix Marin Nov 27 '13 at 10:46