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Algebra 2....

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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.... -

    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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