Question

simplify the following.
(x^-4*z^7)((2x^{2}y)/(z^-1))^-3

Answer 1

Err
\[(x^{-4}z^7)[{(2x^{2}y)\over(z^{-1})}]^{-3}\]
This?

Answer 2

\[ (x^{-4}*z^7)({\frac{2x^{2}y}{z^-1}})^{-3}\]

Answer 3

Yeah, so what I wrote.

Answer 4

yeah.
z to the -1. z^(-1)

Answer 5

So start by distributing the -3 exponent to the numerator & denominator.

Answer 6

i got that. down.
i have
\[(x^{-4})\frac{z^3}{8x^6y^3}\]

Answer 7

oops i mean. (x^-4*z^7)

Answer 8

Good! Now combine the two factors

Answer 9

Into one fractional expression I mean

Answer 10

do i just multiply the (x^-4*z^7) by the numerator and that's all?

Answer 11

And simplify, yes.

Answer 12

ok so i got.
\[\frac{z^{21}}{8xz^{12}x^6y^3}\]

Answer 13

That's not right. I'm not sure how you got that actually? Can you explain?

Answer 14

oops i messed up. should have added the exponents not multiplied. i see my mistake.

Answer 15

When you multiply powers of the same base you do add their exponents.

Answer 16

yeah i figured. thanks. appreciate it.