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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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up vote 3 down vote accepted

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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$$a^6+b^6=(a^3)^2+(b^3)^2=(a^3+b^3)^2-2(ab)^3\text{ and } a^3+b^3=(a+b)^3-3ab(a+b)$$


$$a^6+b^6=(a^2)^3+(b^2)^3=(a^2+b^2)^3-3(ab)^2(a^2+b^2) \text{ and } a^2+b^2=(a+b)^2-2ab$$

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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

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\alpha\beta = -3\,,\quad \alpha + \beta = -2.\quad$ Let $\alpha > 0$ and $\beta < 0.\quad$ Then, $\alpha\verts{\beta} = 3$ and $\verts{\beta} - \alpha= 2.\quad$ Let $\alpha = 2\sinh^{2}\pars{\theta}.\quad$ $\verts{\beta} = 2\cosh^{2}\pars{\theta}$: $$ 3 = \bracks{2\sinh^{2}\pars{\theta}}\bracks{2\cosh^{2}\pars{\theta}} = \sinh^{2}\pars{2\theta}\quad\imp\quad\theta = \half{\rm arcsinh}\pars{\root{3}} $$

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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

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