Compute this determinant only using determinant properties Compute $$\begin{vmatrix}
1 & x & x^2 & x^3 \\
1 & y & y^2 & y^3 \\
1 & z & z^2 & z^3 \\
1 & x+y+z & xy+yz+zx & xyz
\end{vmatrix}$$,where $x,y,z \in \mathbb{R}$.
I tried to create zeroes on the first line,but this doesn't seem to work.
 A: That determinant is $0$, because the fourth column is equal to the sum of:


*

*the first column times $xyz$;

*the second column times $-xy-xz-yz$;

*the third column times $x+y+z$.

A: Here is a solution whose spirit is more "matricial" than "determinantal", but that explains the why and how of this exercice, and gives the result (determinant = 0).
Let us recall first that "the" third degree polynomial with roots $x,y,z$ can be written :
$$(t-x)(t-y)(t-z)=-xyz+(xy+yz+zx)t-(x+y+z)t^2+t^3$$
(we have chosen to use the article "the" because we comply to the classical convention with dominant coefficient is $1$).
In particular, taking $t=x$, the LHS is $0$, thus we have 
$$-xyz+(xy+yz+zx)x-(x+y+z)x^2+x^3=0 \tag{1}$$
Similar identities for $y$ and $z$ :
$$-xyz+(xy+yz+zx)y-(x+y+z)y^2+y^3=0 \tag{2}$$
$$-xyz+(xy+yz+zx)z-(x+y+z)z^2+z^3=0 \tag{3}$$
Now consider the matrix product :
$$\begin{pmatrix}1 & x & x^2 & x^3 \\
1 & y & y^2 & y^3 \\
1 & z & z^2 & z^3 \\
1 & x+y+z & xy+yz+zx & xyz\end{pmatrix}\begin{pmatrix}-xyz\\xy+yz+zx\\-(x+y+z)\\1\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\end{pmatrix} \tag{4}$$
(the three first zeros are due to identities (1), (2), (3) ; the last zero being due to terms cancellations).
What conclusion drawn from (4) ? That the image of a  non zero vector is the zero vector ; as the zero vector is also sent onto the zero vector, the transformation associated with the matrix is not a bijection ; thus its determinant is zero. 
Remark 1 : This is in fact very close to the (direct) solution given by @José Carlos Santos, but for the fact that the coefficients survene in a natural way.
Remark 2 : A similar result ($\det M = 0$) can be generalized to any odd number of parameters (odd in order than the cancellations can be done for obtaining a zero last coefficient). In the even case with letters $x,y,z,t$, we get the following result for the analoguous determinant
$$\det(M)=(t - x)(t - y)(t - z)(y - x)(z - x)(z - y)(t^2x^2 + t^2y^2 + t^2z^2 + x^2y^2 + x^2z^2 + y^2z^2)$$
which is rather simple and a kind of generalization of the Vandermonde determinant.
A: It seems worth giving a complete solution for the general case. Obviously I am indebted to @Jean Marie.
So let $x_1,\dots, x_n$ be variables, and let $a_0,a_1, \dots, a_n$ be the elementary symmetric functions in these, so that
$$
p(X):=\prod_{i}(X-x_i)=\sum_{j=0}^{n} (-1)^{n-j}a_{n-j}X^{j},  
$$ and
$$
a_s=\sum_{i_1<i_2<\dots<i_s} x_{i_{1}}x_{i_{2}}\dots x_{i_{s}}.
$$
Now consider the determinant of the $(n+1)\times (n+1)$ matrix
$$
\begin{bmatrix}
1 & 1 & \dots & 1 & a_0\\
x_1 & x_2 & \dots & x_n & a_1\\
\vdots &\vdots&\ddots &\vdots &\vdots\\
x_1^n & x_2^n & \dots & x_n^n & a_n\\
\end{bmatrix},
$$
whose rows we label for convenience from $0$ to $n$.
As $a_0=1$ the value of the determinant will be unchanged after the following row-operation:
$$
\text{replace   } R_{n}  \text{   by    }\sum_{j=0}^{n} (-1)^{n-j}a_{n-j} R_{j}.
$$
This means we can consider the determinant of this matrix:
$$
\begin{bmatrix}
1 & 1 & \dots & 1 & a_0\\
x_1 & x_2 & \dots & x_n & a_1\\
\vdots &\vdots&\ddots &\vdots &\vdots\\
x_1^{n-1} & x_2^{n-1} & \dots & x_n^{n-1} & a_{n-1}\\
p(x_1) & p(x_2) & \dots & p(x_n) & \sum_{j=0}^{n} (-1)^{n-j}a_{n-j}a_j\\
\end{bmatrix},
$$
which simplifies to
$$
\begin{bmatrix}
1 & 1 & \dots & 1 & a_0\\
x_1 & x_2 & \dots & x_n & a_1\\
\vdots &\vdots&\ddots &\vdots &\vdots\\
x_1^{n-1} & x_2^{n-1} & \dots & x_n^{n-1} & a_{n-1}\\
0 & 0 & \dots & 0 & \sum_{j=0}^{n} (-1)^{n-j}a_{n-j}a_j\\
\end{bmatrix}.
$$
The original determinant is therefore a product of the Vandermonde determinant $\prod_{i<j}(x_i-x_j)$ and the symmetric function $\alpha:=\sum_{j=0}^{n} (-1)^{n-j}a_{n-j}a_j$. It is the latter that we must simplify.
In the case when $n$ is odd $\alpha$ simplifies at once to $0$; the equal terms $a_{n-j}a_j$ and $a_j a_{n-j}$ have opposite sign and so cancel out. 
In the case when $n=2k$ is even things are a little more complicated, but in this case (as we show below) 
$$
\alpha=\sum_{i_1<i_2<\dots<i_k} x_{i_{1}}^2 x_{i_{2}}^2 \dots x_{i_{k}}^2,
$$
the $k$-th elementary symmetric function in the squares of the original variables. 
To see this, note first that 
$$
(-1)^n p(-X)=\prod_{i}(X+x_i)=\sum_{j=0}^{n} a_{n-j}X^{j},  
$$
so that
$$
(\sum_{j=0}^{n} (-1)^{n-j}a_{n-j}X^{j})(\sum_{j=0}^{n} a_{n-j}X^{j})=\prod_{i}(X-x_i) \prod_{i}(X+x_i)=\prod_{i}(X^2 - x_i^2).
$$
If we now pick out the coefficient of $X^{2k}$ on each side we have the promised $
\alpha=\sum_{i_1<i_2<\dots<i_k} x_{i_{1}}^2 x_{i_{2}}^2 \dots x_{i_{k}}^2.
$
