Can anyone help with Simplfying these two equations?

(1/4) ^3/2

and

16 ^-3/2

Question

Can anyone help with Simplfying these two equations?

(1/4) ^3/2 

and

16 ^-3/2

zzr0ck3r · Answer

note: they are not equations
(1/4)^(3/2) = 1^(3/2)/4^(3/2) = 1/4^(3/2) = 1/sqrt(4^3) = 1/sqrt(64) = 1/8 and 1/-8

anonymous · Answer

zz is the second -1/8?

zzr0ck3r · Answer

same thing
$$-\frac{a}{b}=\frac{-a}{b}=\frac{a}{-b}$$

zzr0ck3r · Answer

nono, the first part has two solutions

anonymous · Answer

There is a place for the first answer and a place for the second, what do I write in those places?

zzr0ck3r · Answer

part2
16 ^-3/2
= 1/(16^(3/2))= 1/4^3 and 1/-4^3 
so 1/64 and -1/64

zzr0ck3r · Answer

um the positive one I guess

zzr0ck3r · Answer

1/8

zzr0ck3r · Answer

anonymous · Answer

Ok, I plugged that in so how do I write the second part?

zzr0ck3r · Answer

anonymous · Answer

The second part is 1/64

anonymous · Answer

So the first is 1/8 and the second 1/64?

zzr0ck3r · Answer

yes