$$ x \cdot y = \frac{1}{2 \cdot 2 !} \left( (x + y)^2  (x  y)^2 \right) $$ $$ \begin{eqnarray} x \cdot y \cdot z &=& \frac{1}{2^2 \cdot 3 !} ((x + y + z)^3  (x + y  z)^3 \nonumber \\ && (x  y + z)^3 + (x  y  z)^3 ), \end{eqnarray} $$ $$ \begin{eqnarray} x \cdot y \cdot z \cdot w &=& \frac{1}{2^3 \cdot 4 !} ( (x + y + z + w)^4 \nonumber \\ && (x + y + z  w)^4  (x + y  z + w)^4 \nonumber \\ &+& (x + y  z  w)^4  (x  y + z + w)^4 \nonumber \\ &+& (x  y + z  w)^4 + (x  y  z + w)^4 \nonumber \\ && (x  y  z  w)^4 ). \end{eqnarray} $$ The identity that rewrites a product of $n$ variables $( n \ge 2$, $n \in \boldsymbol{\mathbb{Z}_+})$ as additions of $n$ th power functions is as given below: $$ \begin{eqnarray} & &x_0 \cdots x_1 \cdot x_{n1} = \frac{1}{2^{n  1} \cdot n !} \cdot \sum_{j = 0}^{2^{ n  1} 1} (  1 )^{\sum_{m = 1}^{n  1} \sigma_m(j)} \times ( x_0 + (1)^{\sigma_1(j)} x_1 + \cdots + ( 1)^{\sigma_{n  1}(j)} x_{n  1})^n, \\ & &\sigma_m(j) = r(\left\lfloor \frac{j}{2^{m  1}}\right\rfloor, 2), m (\ge 1), j (\ge 0) \in \mathbb{Z}_+, \; \left\lfloor x \right\rfloor = \max \{ n \in \mathbb{Z}_+ ; n \le x, x \in \mathbb{R} \} \end{eqnarray} $$ where $r(\alpha, \beta)$, $\alpha, \beta (\ge 1) \in \mathbb{Z}_+$ means the remainder of the division of $\alpha$ by $\beta$ such that $r(\alpha, \beta)$ $=$ $\alpha  \beta \left\lfloor \frac{\alpha}{\beta}\right\rfloor$

5Why do you need to recode the subsets of $\left\{1,2,\ldots,n1\right\}$ as numbers in $\left\{0,1,\ldots,2^{n1}1\right\}$ ? – darij grinberg Oct 9 '15 at 2:55

I intend to constructively make an one sided approximation method of a continuous function of many variables by using the general type identity above. Therefore, I summarized the general type identity above as a constructive form. I need to use $\{0,1,...,2^{n1}\}$ type. – Hideaki Okazaki Oct 12 '15 at 0:11
Although not the exactly the same due to $2^{n1}$ instead of $2^n$ terms, the OP's formula seems to be essentially the wellknown polarization formula for homogeneous polynomials, which is stated as following:
Any polynomial $f$, homogeneous of degree $n$ can be written as $f(x)=H(x,\ldots,x)$ for a specific multilinear form $H$. One has the following polarization formula for $H$ (see also this MO post): \begin{equation*} H(x_1,\ldots,x_n) = \frac{1}{2^n n!}\sum_{s \in \{\pm 1\}^n}s_1\ldots s_n f\Bigl(\sum\nolimits_{j=1}^n s_jx_j\Bigr) \end{equation*}
In your case, $f=x^n$, so $H(x_1,\ldots,x_n) = x_1\cdots x_n$ (please note off by 1 indexing).

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@darijgrinberg: thanks! I think I got confused due to the extra $\lfloor . \rfloor$ notation in the OP, and did not feel like verifying :) – Suvrit Oct 9 '15 at 3:22

Thank you very much. I found these identies from n = 2 until n= 6, 22 years ago. Then 3 years ago, I summarized the general type identity above, by using a remainder form. I also proved that general type identity above, by using a recurrence formula. Now I understood that the general type identity above, is a special form of polarization formula in the case of $f =x^n$, and H(x_0, \ldots, x_{n1})$. – Hideaki Okazaki Oct 11 '15 at 23:11

@HideakiOkazaki You are welcome! Very interesting to note that you discovered some of these 22 years ago. I don't know though how old these polarization identities actually are.... – Suvrit Oct 11 '15 at 23:46

1@mostafa You can see this polarization identity for instance in arxiv.org/pdf/math/0504397.pdf  it is mentioned in other places too; I don't know the "oldest / canonical" citation for this though; maybe Alexandrov... – Suvrit Apr 22 '16 at 14:17