Question

I am still learning how to simplify to lowest terms without using negative exponents. I would love some help looking at my answer to this expression.

(b^5/3 c^-1/2)^3

Answer 1

\[(\frac{ b^5 }{ 3c^{-1/2} })^{3}\] We need to cube the top and the bottom, so:
\[\frac{ b^{15} }{ 27c^{-3/2} }\]

Answer 2

Ok, I got the first part right now how do we write the denominator without a negative?

Answer 3

I got c^3/6 how did you get 27?

Answer 4

You need to multiply -1/2 by 3. So you'd get -3/2 as the new exponent. To eliminate the negative, move the term to the numerator to get
\[27b^{15}c^{3/2}\]

Answer 5

Is that c^2/2?

Answer 6

No, it's c^(3/2). Openstudy needs to add the ability to zoom in on equations.

Answer 7

or is that ^1/2 Isn't ^-1/2 x 3 = to c ^3/2

Answer 8

I must be writing the answer incorrectly should it be written as a fraction?

Answer 9

No, theres no fraction, since the negative exponent is in the denominator, it comes up to the numerator. Your answer should be 27(b^15)c^(3/2)

Answer 10

So the correct answer is c^5 b^3/2 not as a fraction but side by side?

Answer 11

Based on the new problem that you sent me, it's c^5 divided by b^(3/2)

Answer 12

This is the original multiplication to simplify:

(b^5/3 c^-1/2)^3

Answer 13

Yes. You triple all the exponents and get b^5 c^(-3/2). The exponent on the c is negative so move it to the bottom. b^5 divided by c^(3/2)