(64n^9)^^{2}/_{3 }(5b^2c^^{1}/_{4})^3

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- May 6th 2012, 04:57 PMjobbles85Simplify equation with steps
(64n^9)^

^{2}/_{3 }(5b^2c^^{1}/_{4})^3 - May 6th 2012, 07:15 PMProve ItRe: Simplify equation with steps
- May 7th 2012, 10:10 AMjobbles85Re: Simplify equation with steps
- May 7th 2012, 10:21 AMskeeterRe: Simplify equation with steps
- May 7th 2012, 12:39 PMjobbles85Re: Simplify equation with steps
(64n^9)^2/3

=642/3n^6

=(3√64^2)(n^6)

=(3√4096)(n^6)

=±16n^6

God I have no clue what I'm doing

(5b^2c^1/4)^3

=(5)^3 (b^2)^3 (c^1/4)^3

=125b^5c^3/4 - May 7th 2012, 12:44 PMskeeterRe: Simplify equation with steps
would have been really easy if you knew that $\displaystyle 64 = 4^3$ ... and it's not $\displaystyle \pm$

last one is correct