lagrange multiplier for more than 2 equality constraints i couldn't do the following question for hours
minimize $\sum_{i=1}^{n}x_{i}^{3}$
s.t.  $\sum_{i=1}^{n}x_{i}=0$
      and $\sum_{i=1}^{n}x_{i}^{2}=n$.
by Lagrange multiplier rule
?
 A: This is a development of my comment, changing the restraints from two to one.
Since $x_{n}=-\sum_{i=1}^{n-1}x_{i}$, the question "find 
$$\min \sum_{i=1}^{n}x_{i}^{3}$$
s.t. 
$$\sum_{i=1}^{n}x_{i}=0$$
and
$$\sum_{i=1}^{n}x_{i}^{2}=n\ "$$
is equivalent to find 
$$\min f(x_{1},x_{2},\ldots ,x_{n-1}),$$
where $f(x_{1},x_{2},\ldots ,x_{n-1})=\left(
\sum_{i=1}^{n-1}x_{i}^{3}\right) -\left( \sum_{i=1}^{n-1}x_{i}\right) ^{3}$
s.t. 
$$c(x_{1},x_{2},\ldots ,x_{n-1})=0,$$
where 
$$c(x_{1},x_{2},\ldots ,x_{n-1})=\left( \sum_{i=1}^{n-1}x_{i}\right)
^{2}-n+\sum_{i=1}^{n-1}x_{i}^{2}=0.$$
The conditions regarding first derivatives are
$$\nabla f(x_{1},x_{2},\ldots ,x_{n-1})-\nabla c(x_{1},x_{2},\ldots
,x_{n-1})\lambda =0\qquad (\ast ),$$
where
$$\nabla f(x)=\begin{pmatrix}\frac{\partial f}{\partial x_{1}} & \ldots  & \frac{\partial f}{\partial x_{n-1}}\end{pmatrix}^{T}$$
$$=%
\begin{pmatrix}
3x_{1}^{2}-3\left( \sum_{i=1}^{n-1}x_{i}\right) ^{2} & \ldots  & 
3x_{n-1}^{2}-3\left( \sum_{i=1}^{n-1}x_{i}\right) ^{2}%
\end{pmatrix}%
^{T}$$
$$\nabla c(x)=%
\begin{pmatrix}
\frac{\partial c}{\partial x_{1}} & \ldots  & \frac{\partial c}{\partial
x_{n-1}}%
\end{pmatrix}%
^{T}$$
$$=%
\begin{pmatrix}
2\left( \sum_{i=1}^{n-1}x_{i}\right) +2x_{1} & \ldots  & 2\left(
\sum_{i=1}^{n-1}x_{i}\right) +2x_{n-1}%
\end{pmatrix}%
^{T}.$$
Condition $(\ast )$ is
$$3x_{1}^{2}-3\left( \sum_{i=1}^{n-1}x_{i}\right) ^{2}-2\lambda \left(
\sum_{i=1}^{n-1}x_{i}\right) -\lambda 2x_{1}=0$$
$$\ldots $$
$$3x_{n-1}^{2}-3\left( \sum_{i=1}^{n-1}x_{i}\right) ^{2}-2\lambda \left(
\sum_{i=1}^{n-1}x_{i}\right) -\lambda 2x_{n-1}=0.$$
All these equations have the same form (the coefficients of $x_k$, $1\lt k\lt n-1$, are independent of $k$). Thus the solution is atained when all the $n-1$ variables are equal
$$x_{1}=\ldots =x_{n-1}=x^{\ast }$$
Hence, we have to solve the new system of two equations
$$3(x^{\ast })^{2}-3\left( \sum_{i=1}^{n-1}x^{\ast }\right) ^{2}-2\lambda
\left( \sum_{i=1}^{n-1}x^{\ast }\right) -2\lambda x^{\ast }=0\qquad (\ast
\ast )$$
$$\left( \sum_{i=1}^{n-1}x^{\ast }\right)
^{2}-n+\sum_{i=1}^{n-1}(x^{\ast })^{2}=0.$$
We obtain, successively
$$3(x^{\ast })^{2}-3n+3\sum_{i=1}^{n}(x^{\ast })^{2}-2\lambda \left(
\sum_{i=1}^{n-1}x^{\ast }\right) -2\lambda x^{\ast }=0$$
$$3(x^{\ast })^{2}-3n+3(x^{\ast })^{2}(n-1)-2\lambda x^{\ast }(n-1)-2\lambda
x^{\ast }=0$$
$$(x^{\ast })^{2}(n-1)-1=0$$
$$x^{\ast }=x_{1}=\ldots =x_{n-1}=\pm \frac{1}{\sqrt{n-1}}.$$
Therefore
$$x_{n}=-\sum_{i=1}^{n-1}x_{i}=\mp \frac{n-1}{\sqrt{n-1}}.$$
Since for $n>2$
$$\left( \frac{1}{\sqrt{n-1}}\right) ^{3}-\left( \frac{n-1}{\sqrt{n-1}}%
\right) ^{3}<-\left( \frac{1}{\sqrt{n-1}}\right) ^{3}+\left( \frac{n-1}{%
\sqrt{n-1}}\right) ^{3},$$
the minimum is obtained at
$$x^{\ast }=x_{1}=\ldots =x_{n-1}=\frac{1}{\sqrt{n-1}},x_{n}=-\frac{n-1}{%
\sqrt{n-1}}$$ 
and its value is equal to
$$\left( \frac{1}{\sqrt{n-1}}\right) ^{3}-\left( \frac{n-1}{\sqrt{n-1}}%
\right) ^{3}.$$

Added: The original problem in $n$ variables is the solution of $n$ equations
$$\nabla f(x_{1},x_{2},\ldots ,x_{n})-\nabla c(x_{1},x_{2},\ldots
,x_{n})\lambda =0\qquad (\ast\ast\ast )$$
plus the two conditions
$$c_{1}(x_{1},x_{2},\ldots ,x_{n})=0$$
$$c_{2}(x_{1},x_{2},\ldots ,x_{n})=0,$$
where
$$\nabla f=%
\begin{pmatrix}
\frac{\partial f}{\partial x_{1}} & \ldots  & \frac{\partial f}{\partial
x_{n}}%
\end{pmatrix}%
^{T}$$
$$\nabla c=%
\begin{pmatrix}
\left( \nabla c_{1}\right) ^{T} & \left( \nabla c_{2}\right) ^{T}%
\end{pmatrix}%
$$
$$\lambda =\begin{pmatrix}
\lambda _{1} & \lambda _{2}\end{pmatrix}^{T}.$$
Note: The equations would be similar for $m$ constraints $c_1,\dots ,c_m.$
Remark: Since I don't know how to write here matrices with more than one row, I wrote them in the transposed form.  
A: Using Lagrange multipliers we get $x_i^2 = \lambda x + \mu$, so that $x_i$ can take only two values, say $X > 0$ and $Y < 0$ (we can't have $X=Y=0$ for obvious reasons). Suppose that $X$ is taken $a$ times, and $Y$ is taken $b = n-a$ times. We have
$aX + bY = 0$
$aX^2 + bY^2 = n$
You can check that the solution is
$X^2 = b/a, Y^2 = a/b$
The objective function that we want to minimize is
$aX^3 + bY^3 = b\sqrt{b/a} - a\sqrt{a/b}$
We want to make the first summand small and the second big. Note that as we increase $a$, the first summand becomes smaller, and the second one becomes bigger, so we might as well choose $a=n-1$ and $b=1$. So $X = 1/\sqrt{(n-1)}$ and $Y = -\sqrt{n-1}$.
Concluding, the optimum is obtained with the sequence $-\sqrt{n-1}, 1/\sqrt{(n-1)} \times (n-1)$, and the optimal value is
$ -(n-1)\sqrt{n-1} + 1/\sqrt{n-1} $
