You may want to review the rules for simplifying exponential expressions first.

Here's a list of the rules covered in those sections, all in one place:

*a*=^{b}a^{c}*a*.^{b+c}- (
*ab*)^{c}=*a*^{c}b^{c}.
- (
*a*)^{b}^{c}=*a*^{bc}.

An example is
(1)^{3}(*a*^{4})(−*a*^{6})^{0}(−2*b*^{2})^{2}.
You need to break it up into pieces. First of all, (1)^{3}
= 1. (This is just arithmetic.) Second,
(−*a*^{6})^{0} is the same as
(−1*a*^{6})^{0}, which is
(−1)^{0}(*a*^{6})^{0}, or
1*a*^{0}. Since multiplying by 1 doesn't do
anything, this is really the same thing
as *a*^{0}. Finally,
(−2*b*^{2})^{2} simplifies to
4*b*^{4}. (This is the problem in the last
paragraph, only with *b* instead of *x*.) So altogether
we have
1*a*^{4}*a*^{0}(4*b*^{4}). Now,
1 * 4 = 4, and by the first
rule *a*^{4}*a*^{0}
= *a*^{4}, so our final answer is
4*a*^{4}*b*^{4}. (We can't
combine *a*^{4} and *b*^{4},
since *a* and *b* are different letters.)

There are two shortcuts you can use. First, anything raised to the 0 power is 1. Second, anything raised to the 1 power is itself.