This question was a little misleading because the book gives you the truth table right below the question. All you have to do is compare the fifth and sixth columns to see they are the same. Also, see that columns seven and eight are the same for the second equation.
There are many ways to do this. Here's one. The original equation is: \[ E = (AB + AC + BC)(\overline{ABC}) \] Now use DeMorgan's to rewrite the last factor: \[ E = (AB + AC + BC)(\bar{A}+\bar{B}+\bar{C}) \] Distribute the last product across each term of the first sum: \[ E = AB(\bar{A}+\bar{B}+\bar{C}) + AC(\bar{A}+\bar{B}+\bar{C}) + BC(\bar{A}+\bar{B}+\bar{C}) \] Now distribute each term above. I'll do the first one: \[ AB(\bar{A}+\bar{B}+\bar{C}) = AB\bar{A}+AB\bar{B}+AB\bar{C} = 0 + 0 + AB\bar{C} = AB\bar{C} \] If we do this for each term we end up with: \[ E = AB\bar{C}+A\bar{B}C+\bar{A}BC \] which is what we wanted.
If a circuit has $n$ inputs then each value can take on one of two values, 0 or 1, which is $2^n$ possible different inputs ranging from all zeroes $\underbrace{00\ldots0}_{n}$ to all ones $\underbrace{11\ldots1}_{n}$. That is, in base ten from $0$ to $2^n-1$.
Look at class notes. We did this in class.