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How to express $$a^7+b^7+c^7$$ in terms of symmetric polynomials ${\sigma}_{1}=a+b+c$, ${\sigma}_{2}=ab+bc+ca$ and ${\sigma}_{3}=abc$ ?

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  • $\begingroup$ Newton's identities $\endgroup$ Feb 21 '15 at 14:34
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    $\begingroup$ But Newton's identity will be a very tedious process especially for my question and for higher powers $\endgroup$
    – user196761
    Feb 21 '15 at 14:39
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    $\begingroup$ Not so very tedious, see my answer. Though I would program this for higher powers. $\endgroup$ Feb 21 '15 at 15:10
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We can take a slightly different approach. We have that $a,b,c$ are the roots of the polynomial: $$ p(x)=x^3-\sigma_1 x^2+\sigma_2 x-\sigma_3 $$ hence the eigenvalues of the companion matrix: $$ M = \left(\begin{array}{ccc} 0 & 0 & \sigma_3 \\ 1 & 0 & -\sigma_2 \\ 0 & 1 & \sigma_1\end{array}\right) $$ so: $$\begin{eqnarray*} a^7+b^7+c^7 &=& \operatorname{Tr}\left(M^7\right)\\&=&\sigma_1^7 - 7 \sigma_1^5 \sigma_2 + 14 \sigma_1^3 \sigma_2^2 - 7 \sigma_1 \sigma_2^3 + 7 \sigma_1^4 \sigma_3 - 21 \sigma_1^2 \sigma_2 \sigma_3 + 7 \sigma_2^2 \sigma_3 + 7 \sigma_1 \sigma_3^2.\end{eqnarray*} $$

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Hint:

$$a^n+b^n+c^n=(a+b+c)(a^{n-1}+b^{n-1}+c^{n-1})-(ab+bc+ca)(a^{n-2}+b^{n-2}+c^{n-2})+abc(a^{n-3}+b^{n-3}+c^{n-3}), \forall n\in\mathbb N_{\ge 3}$$

$$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$$

Using this you can express $a^n+b^n+c^n, \forall n\in\mathbb N$ in terms of symmetric polynomials ${\sigma}_{1}=a+b+c$, ${\sigma}_{2}=ab+bc+ca$ and ${\sigma}_{3}=abc$.

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Calling $\sigma_1=X$, $\sigma_2=Y$ and $\sigma_3=Z$ for readability, Newton's identities give for the power sums $p_1$ in $3$ variables: $$\begin{align} p_1&=X,\\ p_2&=Xp_1-2Y &&=X^2-2Y,\\ p_3&=Xp_2-Yp_1+3Z&&=X^3-3XY+3Z,\\ p_4&=Xp_3-Yp_2+Zp_1&&=X^4-4X^2Y+4XZ+2Y^2,\\ p_5&=Xp_4-Yp_3+Zp_2&&=X^5-5X^3Y+5X^2Z+5XY^2-5YZ,\\ p_6&=Xp_5-Yp_4+Zp_3&&=X^6-6X^4Y+6X^3Z+9X^2Y^2-12XYZ-2Y^3+3Z^2,\\ p_7&=Xp_6-Yp_5+Zp_4\\ &\qquad=\rlap{X^7-7X^5Y+7X^4Z+14X^3Y^2-21X^2YZ-7XY^3+7XZ^2+7Y^2Z}.\\ \end{align} $$ Slightly boring, but not really difficult to compute. Actually I think the computation is very close to what happens in computing the powers $M^k$ for the matrix $M$ in the answer of Jack D'Aurizio. Note that there one only needs to compute the last column of each matrix, since every column of $M^k$ except the last is copied from the column one place further in $M^{k-1}$.

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Hint: Expand $(a+b+c)^7=\sigma_1^7$ with Newton to find the terms you need to subtract from it to get $a^7+b^7+c^7.$ I think it's not too high of a power to do that, since its expansion has $\frac{8\cdot9}{2}=36$ terms, arguably not too many, but after all one can also compute $(a+b+c)^6$ by squaring $(a+b+c)^3$ and then multiply by $a+b+c=\sigma_1$.

For example, for the square we have $$\require\color(a+b+c)^2=\sigma_1^2=a^2+b^2+c^2+2ab+2ac+2bc =a^2+b^2+c^2+2(ab+bc+ca) \iff \ \color\red{a^2+b^2+c^2=\sigma_1^2-2\sigma_2}$$

and for the cube $$\require\color (a+b+c)^3=\sigma_1^3=a^3+b^3+c^3+3ab^2+3ac^2+3ba^2+3bc^2+3ca^2+3cb^2+6abc=\\ a^3+b^3+c^3+3\left(ab(a+b)+bc(b+c)+ca(c+a)\right)+6abc=\\ a^3+b^3+c^3+3(a+b+c)(ab+bc+ca)-9abc+6abc \iff \\ \color\red{a^3+b^3+c^3 = \sigma_1^3-3\sigma_1\sigma_2+3\sigma_3}.$$

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  • $\begingroup$ It is $a^7+b^7+c^7$ rather than $(a+b+c)^7=\sigma_1^7$. $\endgroup$ Feb 21 '15 at 14:37
  • $\begingroup$ @MarcvanLeeuwen I know, see the edit. $\endgroup$ Feb 21 '15 at 16:25
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Let $a,b,c$ be the roots of $p(x)=x^3-\sigma_1x^2+\sigma_2x-\sigma _3=0$ and let $s_r=a^r+b^r+c^r$

Then $0=a^rp(a)+b^rp(b)+c^rp(c)=s_{r+3}-\sigma_1s_{r+2}+\sigma_2s_{r+1}-\sigma_3s_r$ gives a recurrence for $s_r$.

Start with $s_0=3, s_1=\sigma_1, s_2=\sigma_1^2-2\sigma_2$

You need $abc\neq 0$ - but if one of the roots is zero you can divide through by $x$ and do something equivalent with the resulting quadratic. The zero adds nothing to any of the sums.

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