Probability that on three rolls of dice, there will be at least one 6 showing up? What is the probability that on three rolls of dice, there will be at least one 6 showing up?
Attempt: 
Since there can be one six or two sixes or three sixes on three rolls, I considered separate cases and added them up.
So $(1/6)(5/6)(5/6) + (1/6)(1/6)(5/6) + (1/6)(1/6)(1/6) = 31/216$, but answer is incorrect as per book.  Can anyone suggest where I am wrong ?

You have forgotten to include th fact that any six can occur in any of the rolls, i.e. you have multiply the probabilities for exactly one and two sixs by three, giving
$$P=3\cdot\frac{1}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}+3\cdot\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{5}{6}+\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}=\frac{91}{216}.$$
It may be easier to compute the complementary probability of no six showing up: $$P^c=\left(\frac{5}{6}\right)^3=\frac{125}{216}$$
and using $$P=1-P^c=\frac{216-125}{216}=\frac{91}{216}$$
