A faster way of calculating this determinant? I'm doing a problem involving Cramer's rule, and one of the determinants I have to work with is as follows:
\begin{vmatrix}
1&1&1\\
a&b&c\\
a^3&b^3&c^3
\end{vmatrix}
So I started off by getting the matrix to a triangular matrix so I can just take the product of the diagonal cells by doing this:
\begin{equation}
R_1 \times-c + R_2
\end{equation}
\begin{equation}
R_3 \times -\frac{1}{c^3} + R_1
\end{equation}
\begin{equation}
R_2 \times(\frac{1-\frac{b^3}{c^3}}{b-c})+R_1
\end{equation}
I then got this matrix
\begin{vmatrix}
((1-\frac{a^3}{c^3})-(a-c)(\frac{1-\frac{b^3}{c^3}}{b-c}))&0&0\\
a-c&b-c&0\\
a^3&b^3&c^3\\
\end{vmatrix}
By summing the diagonal cells, I got this as a final answer:
\begin{equation}
(c-a)(a-b)(b-c)(a+b+c)
\end{equation}
However, it was a lengthy process, and I can't help but thinking this is not the type of calculations I can afford the time in a written exam, especially if I have to repeat this four times solving three linear equations with Cramer's rule. 
I'd really appreciate it if you have any thoughts as to speed this process up. 
Just as an added detail, the original question was (translated from Chinese):
With regards to the following set of equations
\begin{equation}
x + y + z = 1\\
ax + by + cz = d\\
a^3x + b^3y + c^3z = d^3
\end{equation}


*

*Under what conditions can Cramer's rule be used?

*Please solve the set of equations with Cramer's rule. 
For the first question, the immediate thought that pops up is that
\begin{equation}
det(A) ≠ 0
\end{equation}
For the second question, it's simply
\begin{equation}
\frac{\begin{vmatrix}A_1\end{vmatrix}}{\begin{vmatrix}A\end{vmatrix}},
\frac{\begin{vmatrix}A_2\end{vmatrix}}{\begin{vmatrix}A\end{vmatrix}},
\frac{\begin{vmatrix}A_3\end{vmatrix}}{\begin{vmatrix}A\end{vmatrix}} 
\end{equation}
which is the part I'm having trouble solving quickly. 
Again, appreciate any hints or thoughts on this. Thanks. 

You can compute the determinant of a generic $3\times 3$ matrix using a neat trick, if we have:
$$\mathbf{A}=\begin{pmatrix}a & b & c \\ d &  e & f \\ g & h & i \end{pmatrix}$$
Then we have the sum of the diagonals (highlighted in green) minus the sum of the antidiagonals (highlighted in red) as follows:
$$\det(\mathbf{A}) = \underbrace{aei}_{\begin{pmatrix}{\color{green} a} & b & c \\ d & {\color{green} e} & f \\ g & h & {\color{green} i}\end{pmatrix}} + \underbrace{bfg}_{\begin{pmatrix}a & {\color{green}b} & c \\ d & e & {\color{green} f} \\ {\color{green} g} & h & i\end{pmatrix}} + \underbrace{cdh}_{\begin{pmatrix}a & b & {\color{green} c} \\ {\color{green} d} & e & f \\ g & {\color{green} h} & i\end{pmatrix}} - \underbrace{afh}_{\begin{pmatrix}{\color{red} a} & b & c \\ d & e & {\color{red} f} \\ g & \color{red} h & i\end{pmatrix}} - \underbrace{bdi}_{\begin{pmatrix}a & \color{red} b & c \\ \color{red} d & e & f \\ g & h & \color{red} i\end{pmatrix}} - \underbrace{ceg}_{\begin{pmatrix}a & b & \color{red}c \\ d & \color{red} e & f \\ \color{red} g & h & i\end{pmatrix}}$$
Thus if we have your matrix:
$$\mathbf{M}=\begin{pmatrix}1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3}\end{pmatrix}$$
Then:
$$\det(\mathbf{M})=bc^{3}+ca^{3}+ab^{3}-cb^{3}-ac^{3}-ba^{3}$$
