Posted by Anonymous on .
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,
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