Game Development Reference
In-Depth Information
n
This problem combines a cube root with the multiplicative inverse of a value:
1
8
3
¼
1
8
1
2
8
3
3
p
¼
¼
In this instance, the negative exponent (
1/3) calls for the multiplicative
inverse of the cube root of 8. Having expressed the problem in this form, you
can then extract the root and arrive at 1/2.
n
This problem involves working with both a negative exponent and the need
to simplify an expression that involves a root and a power:
1
4
2
¼
1
ð
1
2
3
¼
1
8
4
2
¼
p
Þ
3
¼
With this problem, you begin by dealing with the negative exponent. You
change the exponent into radical notation and then work on simplifying the
radical. There are a few approaches to this. One involves emphasizing that
p
¼
2. You can then raise 2 to the third power, which results in 8. An
alternative approach is to write the denominator as
ð
4
Þ
q
p
4
4
4
p
64
3
¼
¼
¼
8
n
This problem involves a situation in which you must carry out a multi-
plication operation in the denominator of the fraction.
4
3
4
3
4
3
¼
4
3
3
¼
4
0
4
3
¼
¼
1
4
3
To carry out the multiplication, since you are working with the same base
(4), you can add the exponents (2/3
þ
5/3) of the two numbers in the
denominators. Then, to carry out the division, you can subtract the ex-
ponent of the denominator from the exponent of the numerator. The result
is 0, and any number raised to the power of 0 is equal to 1.
n
This problem has a denominator that includes several radical expressions:
25
3
25
3
1
1
25
3
þ
3
1
25
1
1
25
8
1
25
2
¼
1
50
3
p
¼
8
3
¼
8
3
¼
8
3
¼
8
3
¼
25
3
25
3
Resolution of the problem involves first moving the numerator to the de-
nominator, and then adding the two exponents of 25. When you add 1/3 and
2/3, the result is 1, and any number raised to the power of 1 is the number
