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(3x^4y-5)^3 ??
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3x^12 y^3 125
Is that how it's solved?
thats the final answer, how it is solved it by multiplying all the exponents by 3
not its not
yes it is.
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okay thank you very much (: do you know how to solve them as well with a root number with exponent?
(3x^4y-5)^3 = (3x^4y-5)(3x^4y-5)(3x^4y-5) =
oh wait...crap you rirght!
so you are telling me (2+4)^2 = 2^2 + 4^2
yes I am
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i had a brain malfunction
how would I solve for it?
like rock said
(a+b)^3 = = a^3+3a^2b+3ab^2+b^3
for you a = 3x^4y and b = -5
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alrighty thank you(:
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