Find determinant Find determinant of A.
$${det}A =\left( \begin{array}{cc}
1 & 1 & 1 & 1 & 1\\
a & b & c & d & x\\
a^2 & b^2 & c^2 & d^2 & x^2\\
a^3 & b^3 & c^3 & d^3 & x^3\\
a^4 & b^4 & c^4 & d^4 & x^4\end{array} \right)$$
I was thinking that I should reduce ones on top, but then I would be stuck with this:
$${det}A =\left( \begin{array}{cc}
b-a & c-a & d-a & x-a\\
b^2-a^2 & c^2-a^2 & d^2-a^2 & x^2-a^2\\
b^3-a^3 & c^3-a^3 & d^3-a^3 & x^3-a^3\\
b^4-a^4 & c^4-a^4 & d^4-a^4 & x^4-a^4\end{array} \right)$$
Any suggestions?

One of the tricks to solving this is using the fact that $\text{det }A^T = \text{det }A$.
Use this once to get
$\text{det }A = \begin{vmatrix}1&1&1&1&1 \\ a&b&c&d&x \\ a^2&b^2&c^2&d^2&x^2 \\ a^3&b^3&c^3&d^3&x^3 \\ a^4&b^4&c^4&d^4&x^4 \end{vmatrix} = \begin{vmatrix}1&a&a^2&a^3&a^4 \\ 1&b&b^2&b^3&b^4 \\ 1&c&c^2&c^3&c^4 \\ 1&d&d^2&d^3&d^4 \\ 1&x&x^2&x^3&x^4 \end{vmatrix}$
Reduce this to get $\text{det }A = \begin{vmatrix}1&a&a^2&a^3&a^4 \\ 0&b-a&b^2-a^2&b^3-a^3&b^4-a^4 \\ 0&c-a&c^2-a^2&c^3-a^3&c^4-a^4 \\ 0&d-a&d^2-a^2&d^3-a^3&d^4-a^4 \\ 0&x-a&x^2-a^2&x^3-a^3&x^4-a^4 \end{vmatrix}$
Now use the initial property with transposes again to get $\text{det }A = \begin{vmatrix}1&0&0&0&0 \\ a&b-a&c-a&d-a&x-a \\ a^2&b^2-a^2&c^2-a^2&d^2-a^2&x^2-a^2 \\ a^3&b^3-a^3&c^3-a^3&d^3-a^3&x^3-a^3 \\ a^4&b^4-a^4&c^4-a^4&d^4-a^4&x^4-a^4 \end{vmatrix}$
Row reduce yet again to get $\text{det }A = \begin{vmatrix}1&0&0&0&0 \\ 0&b-a&c-a&d-a&x-a \\ 0&b(b-a)&c(c-a)&d(d-a)&x(x-a) \\ 0&b^2(b-a)&c^2(c-a)&d^2(d-a)&x^2(x-a) \\ 0&b^3(b-a)&c^3(c-a)&d^3(d-a)&x^3(x-a) \end{vmatrix}$
And expand to get $\text{det }A = \begin{vmatrix} b-a&c-a&d-a&x-a \\ b(b-a)&c(c-a)&d(d-a)&x(x-a) \\ b^2(b-a)&c^2(c-a)&d^2(d-a)&x^2(x-a) \\ b^3(b-a)&c^3(c-a)&d^3(d-a)&x^3(x-a) \end{vmatrix}$
Now, since we know that $\text{det }A^T = \text{det A}$ and that the row operation of multiplying by a scalar multiplies the determinant by that scalar, we can use "column operations" to reduce this to
$$
\text{det} A = (b-a)(c-a)(d-a)(x-a) \begin{vmatrix} 1&1&1&1 \\ b&c&d&x \\ b^2&c^2&d^2&x^2 \\ b^3&c^3&d^3&x^3 \end{vmatrix}
$$
This matrix looks a lot like the beginning matrix, and you're going to solve it in basically the exact same way so that your answer comes out to
$$
\text{det} A = (b-a)(c-a)(d-a)(x-a)(c-b)(d-b)(x-b)(d-c)(x-c)(x-d)
$$.
On another note, this generalizes to what others were talking about in the comments, the Vandermonde matrix (or the transpose of the Vandermonde matrix in your case, but they have the same determinant). If we define the Vandermonde determinant $\Delta$ to be
$$
\Delta = \begin{vmatrix} 1&1&\cdots&1 \\ x_1&x_2&\cdots&x_n \\ x_1^2&x_2^2&\cdots&x_n^2 \\ \vdots&\vdots&\ &\vdots \\x_1^{n-1}&x_2^{n-1}&\cdots&x_n^{n-1} \end{vmatrix}
$$
Then $$\Delta = \prod_{1 \leq i < j \leq n} {x_j-x_i}$$
You could use this result to quickly do your problem, but you should at least see how one would go about proving this result.
