It's just

    ax1^3 + bx1^2 + cx1 + d = y1
    ax2^3 + bx2^2 + cx2 + d = y2
    ax3^3 + bx3^2 + cx3 + d = y3
    ax4^3 + bx4^2 + cx4 + d = y4

As a matrix:

    [x1^3 + x1^2 + x1 + 1] [a] = [y1]
    [x2^3 + x2^2 + x2 + 1] [b] = [y2]
    [x3^3 + x3^2 + x3 + 1] [c] = [y3]
    [x4^3 + x4^2 + x4 + 1] [d] = [y4]

If the matrix is M, we just want to find M^(-1)y.

I think hackers should know enough linear algebra to do that, without "having to go asking the maths gods".