Is (a-3) a factor of (a^3-6a^2+29) ?

If (a - 3) is a factor then (a - 3) will be 0 when a = 3. If you put a = 3 into the given cubic, you should get 0. Doing that, does not give 0.

f (3) = 8 - 6*9 + 29

f (3) = 8 - 54 + 29

f (3) = - 17

As further "proof", here is the graph. Notice that the graph cuts the x axis around - 2, 2.5 and 4.2 (These are rough guesses). The factors are

(a+2) (a - 2.5) (a - 4.2)