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Simplifying [(8 b^9 c^5)^(1/3)] / [27 b^(-3) c^2]
- Thread starter RicanGurl
- Start date
Review all your power rules:
\(\displaystyle a^{m}a^{n} = a^{m+n}\)
\(\displaystyle \left(a^{m}\right)^{n} = a^{m \cdot n}\)
\(\displaystyle \frac{a^{m}}{a^{n}} = a^{m-n}\)
\(\displaystyle a^{-m} = \frac{1}{a^{m}}\)
These are the ones that apply to this problem. If you don't quite understand, then I think you need some reviewing to do. You can click here for a review/tutorial.
\(\displaystyle a^{m}a^{n} = a^{m+n}\)
\(\displaystyle \left(a^{m}\right)^{n} = a^{m \cdot n}\)
\(\displaystyle \frac{a^{m}}{a^{n}} = a^{m-n}\)
\(\displaystyle a^{-m} = \frac{1}{a^{m}}\)
These are the ones that apply to this problem. If you don't quite understand, then I think you need some reviewing to do. You can click here for a review/tutorial.
Not quite right. Do you know BEDMAS? It's an acronym for what operations you should perform FIRST and in ORDER.
B - brackets
E - exponents
D - division
M - multiplicadtion
A - addition
S - subtraction
(Well it doesn't matter what the order D and M are as well as A and S).
Anyway, you start with the expression with the brackets & exponents first, which is: \(\displaystyle \left(8b^{9}c^{5}\right)^{\frac{1}{3}}\). What power rule can you use to simplify this?
B - brackets
E - exponents
D - division
M - multiplicadtion
A - addition
S - subtraction
(Well it doesn't matter what the order D and M are as well as A and S).
Anyway, you start with the expression with the brackets & exponents first, which is: \(\displaystyle \left(8b^{9}c^{5}\right)^{\frac{1}{3}}\). What power rule can you use to simplify this?
Re: i have
Hello, RicanGurl!
You've already broken several rules.
You may need more help than we can give . . .
We have: \(\displaystyle \L\:\frac{(8b^9c^5)^{\frac{1}{3}}}{27b^{-3}c^2}\)
. . \(\displaystyle \L=\;\frac{(8)^{\frac{1}{3}}(b^9)^{\frac{1}{3}}(c^5)^{\frac{1}{3}}}{27b^{-3}c^2}\)
. . \(\displaystyle \L=\:\frac{2\cdot b^3\cdot c^{\frac{5}{3}}}{27\cdot b^{-3}\cdot c^2}\)
Can you finish it?
Hello, RicanGurl!
I have done this so far
(8b12c3) 1/3
----------
(27)
You've already broken several rules.
You may need more help than we can give . . .
We have: \(\displaystyle \L\:\frac{(8b^9c^5)^{\frac{1}{3}}}{27b^{-3}c^2}\)
. . \(\displaystyle \L=\;\frac{(8)^{\frac{1}{3}}(b^9)^{\frac{1}{3}}(c^5)^{\frac{1}{3}}}{27b^{-3}c^2}\)
. . \(\displaystyle \L=\:\frac{2\cdot b^3\cdot c^{\frac{5}{3}}}{27\cdot b^{-3}\cdot c^2}\)
Can you finish it?