Homework Help: Algebra 2....

Posted by Anonymous on Sunday, March 10, 2013 at 10:21pm.

Simplify.(Is this problem more clear than the previous post?)
First part..
a^4/3 times c^1/6 all over 9b^3...everything is in the parenthesis and is raised to the power of -2
MULTIPLY TO THIS SECOND SET:
a^3 times b^2 all over 3c.. everything is in the parenthesis and is raised to the power of -1/4

Algebra 2.... - Steve, Monday, March 11, 2013 at 1:37am
If you have

[(a^4/3 * c^1/6) / (9b^3)]^-2
That's (81b^6) / (a^8/3 * c^1/3)

[(a^3 * b^2)/(3c)]^-1/4 is
(3^1/4 * c*1/4) / (a^3/4 * b^1/2)

multiply to get

(81*3^1/4 * b^11/2) / (a^41/12 * c^1/12)

messy

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