81r^3 - 3t^3 = 0 has solutions when (t/r)^3 = 27, so t/r = 3, 3 omega or 3 omega^2, where omega is the complex cube root of 1, so (3r - t), (3 omega r - t) and (3 omega ^2 r - t) are all factors of 81r^3 - 3t.
If we restrict ourselves to real coefficients, the only linear factor is (3r - t). The other (quadratic) factor will be of the form (ar^2 + brt + ct^2). We could calculate this using omega, but let's stick to real numbers for simplicity...
Let's try multiplying it out:
(3r - t)(ar^2 + brt + ct^2) = 3ar^3 + (3b - a)r^2t + (3c - b)rt^2 -ct^3
Comparing the coefficients with the original polynomial, we get:
3a = 81
3b - a = 0
3c - b = 0
c = 3
Hence a = 27, b = 9 and c = 3.
So 81r^3 - 3t^3 = (3r - t)(27r^2 + 9rt + 3t^2) = 3(3r - t)(9r^2 + 3rt + t^2).
