#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)#.
