For which $a$, $b$, $c$ does this linear system have exactly one solution? $x+ay+a^2z=0$, $x+by+b^2z=0$, $x+cy+c^2z=0$ 
For which $a$, $b$, $c$ does this linear system have exactly one solution?
$$x + ay + a^2z = 0$$
$$x + by + b^2z = 0$$
$$x + cy + c^2z = 0$$

I started this problem by recognizing that if the RREF of a linear system's augmented matrix has a leading 1 in every column except the last, then the system must have exactly one solution. The system's augmented matrix reduces to
$$
    \begin{matrix}
    1 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    \end{matrix}
$$
which indicates that no matter the values of $a, b, c$, $x = 0, y = 0,$ and $z = 0$. But one can see that if $a = 0$ and $b = 0$, then $x = 0$ and the third equation of the system becomes the equation of a line, indicating an infinite number of solutions, meaning that no two of $a,b,c$ van be equal to $0$.

So my best guess is that the system has exactly one solution when $a,b,c \in \mathbb{R}$ and $a, b \neq 0$. Is this correct? Close at all? How would I go about proving this?

 A: HINT
According to Cramer's rule, this system admits a unique solution iff the determinant of the coefficient matrix is different from zero. Precisely,
\begin{align*}
\begin{vmatrix}
1 & a & a^{2}\\ 
1 & b & b^{2}\\ 
1 & c & c^2\\ 
\end{vmatrix}\neq 0
\end{align*}
Such matrix is widely known as the Vandermonte's matrix, and its determinant equals
\begin{align*}
\begin{vmatrix}
1 & a & a^{2}\\ 
1 & b & b^{2}\\ 
1 & c & c^2\\ 
\end{vmatrix} = (c-b)(c-a)(b-a)
\end{align*}
Can you take it from here?
A: The other solution uses determinants which is going to be the standard route to solving this problem. You mentioned RREF which I don't think is impossible, but you should be careful with putting it in that form:
$$\begin{pmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{pmatrix}$$
subtracting row 1 from row 2 and row 1 from row 3:
$$\begin{pmatrix}1&a&a^2\\0&b-a&b^2-a^2\\0&c-a&c^2-a^2\end{pmatrix}$$
Now we immediately see that if $b=a$ or $a=c$ then RREF will have at least one row which is all 0's. What does that mean about the number of solutions?
If not then we can divide by $b-a$ and $c-a$:
$$\begin{pmatrix}1&a&a^2\\0&1&b+a\\0&1&c+a\end{pmatrix}$$
Subtracting $a$ times row 2 from row 1 and row 2 from row 3:
$$\begin{pmatrix}1&0&-ba\\0&1&b+a\\0&0&c-b\end{pmatrix}$$
We again notice that if $b=c$ then there is a row of all 0s. If not then we can divide row 3 by $c-b$ to get:
$$\begin{pmatrix}1&0&-ba\\0&1&b+a\\0&0&1\end{pmatrix}$$
now adding $ba$ times row 3 to row 1 and subtracting $b+a$ times row 3 from row 2:
$$\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$$
What does this mean?
