Question

Simplify (7x^-3*y^2)^-3

Answer 1

First we know that a negative exponent means we divide the whole thing by 1, which gives\[\frac{ 1 }{ (7x^{-3}y^2)^3 }\] Then we look at inside piece, gives \[\frac{ 1 }{ (\frac{y^2}{7x^{3}})^3 }\] which simplifies more gives \[\frac{x^9}{343y^6}\]

Answer 2

Remember that raising to a negative power means inverting it as a fraction and then raising it to the positive power:
\[(\frac{y^{2}}{7x^3})^{-3}\]\[(\frac{7x^3}{y^{2}})^{3}\]\[\frac{7^{3}x^{9}}{y^{6}}\]

Answer 3

The approach @absolutemaster takes is correct also (he just does things in a different order), except he accidentally put the 7^3 (343) next to the y at the end instead of next to the x.

Answer 4

@ChristianGeek Nop, the 7 in your first step should be beside 7^2 and not x^3

Answer 5

@absolutemaster Sorry, it's in the right place, just as it is in yours in the second to last step. You moved it when you inverted the final fraction.

Answer 6

Wait a minute...sorry, you're correct. Your formulas aren't, but your final answer is. My bad.

Answer 7

Oh yea true about the second step, my 7 should be beside y^2

:)