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I recently learnt out about this formula:

$$a^3 + b^3 + c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - ac - bc)$$

Is there a way of generalizing for powers other than $3$, i.e.

$$a^n + b^n + c^n + \mathop{???} = \mathop{???} $$

where the RHS is in terms of $(a^{n-1} + b^{n-1} + c^{n-1})$, $(a^{n-2} + b^{n-2} + c^{n-2})$, ... , $(a^{1} + b^{1} + c^{1})$, just like the formula for $n=3$.

Sangchul Lee
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physBa
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2 Answers2

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Too long for a comment:

One possible generalization is the circulant determinant, which tells that

$$ \det \begin{pmatrix} a_1 & a_2 & a_3 & \cdots & a_n \\ a_n & a_1 & a_2 & \cdots & a_{n-1} \\ a_{n-1} & a_n & a_1 & \cdots & a_{n-2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_2 & a_3 & a_4 & \cdots & a_1 \end{pmatrix} = \prod_{\omega \mathop{:} \omega^n = 1} (a_1 + a_2 \omega + a_3 \omega^2 + \cdot + a_n \omega^{n-1}). $$

When $n = 2$, this gives

$$ a^2 - b^2 = \det\begin{pmatrix} a & b \\ b & a \end{pmatrix} = (a + b)(a - b). $$

When $n = 3$, then with any $\omega$ satisfying $\omega^2 + \omega + 1 = 0$,

\begin{align*} a^3 + b^3 + c^3 - 3abc &= \det\begin{pmatrix} a & b & c \\ c & a & b \\ b & c & a \end{pmatrix} \\ &= (a + b + c)(a + b\omega + c\omega^2)(a + b\omega^2 + c\omega) \\ &= (a + b + c)(a^2 + b^2 + c^2 + (\omega + \omega^2)(ab + bc + ca)) \\ &= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca). \end{align*}

Unfortunately, the formula quickly goes out of control even starting at $n = 4$:

\begin{gather*} a^4 - b^4 + c^4 - d^4 - 2a^2c^2 + 2b^2d^2 - 4da^2b + 4ab^2c - 4bc^2d + 4cd^2a\\ = (a+b+c+d)(a-b+c-d)(a^2 + b^2 + c^2 + d^2 - 2ac - 2bd) \end{gather*}

Sangchul Lee
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1

For cases of $n=5$ and $n=7$, let's have a look at the following identities

$$(x+y+z)^3 - (x^3+y^3+z^3) = 3(x+y)(x+z)(y+z)$$

$$(x+y+z)^5 - (x^5+y^5+z^5) = 5(x+y)(x+z)(y+z)(x^2+y^2+z^2+xy+xz+yz)$$

$$(x+y+z)^7 - (x^7+y^7+z^7) = 7(x+y)(x+z)(y+z)((x^2+y^2+z^2+xy+xz+yz)^2+xyz(x+y+z))$$

The three identities listed above are called Lame-type identities (see here). Algebraically manipulating the second and third identities give;

Let $$A = x+y+z,$$ $$B = x^2+y^2+z^2,$$ $$C = x^3+y^3+z^3,$$ $$E = x^5+y^5+z^5,$$ $$G = x^7+y^7+z^7,$$ then
\begin{equation} 6E=A^5-5BA^3+5CA^2+5BC \tag{1} \end{equation}

\begin{equation} 36G=A^7+7CA^4-21B^2A^3+28C^2A+21B^2C \tag{2} \end{equation}

By the way, I think they have nice applications. For example, if the values of $A, $ $B, $ and $E$ are given, $x $, $y $ and $z$ can be found with ease since $C$ can be generated from (1) and that, solving $A, $ $B, $ and $C$ simultaneously won't be difficult. Same thing goes for $A, $ $B, $ and $G$.

Shuaib Lateef
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