Question

I just need someone to come behind me and check my work, and maybe help if needed.

\[\left( \frac{ x ^{-6}y ^{3} }{ x ^{-3}y ^{3}} \right)^{1/3}\]

Answer 1

\[\left( \frac{ x ^{-2}y }{ x ^{-1}y } \right)\]

Answer 2

keep going

Answer 3

\[x ^{-2 - (-1)} = x ^{-3}\]

Answer 4

So now would it be \[\frac{ y }{ x ^{3}}\]

Answer 5

You have made a mistake, the best way to deal with these questions is to bring all the variables up from the denominator using the rule,

\[x^{-1} = \frac{1}{x}\]

Answer 6

Then just add all the exponents on the same variables

Answer 7

So \[\frac{ 3 }{ x } ?\]

Answer 8

for example:

\[\frac{xy^{2}}{z^{2}x^{-2}} = x^{1}z^{-2}y^{2}x^{2}\]

Answer 9

now it is easy to add variables,
for x

1 + 2 = 3
so we have x^(3)
so we have

(x^1)z^-2(y^2) = ((x^1)(y^2))/z^-2

Answer 10

oops

(x^3)z^-2(y^2) = ((x^3)(y^2))/z^-2

Answer 11

made a mistake

Answer 12

I'm sorry, I don't really get this principle. What does that last equation you posted mean?

Answer 13

wow I keep making mistakes

(x^3)z^-2(y^2) = ((x^3)(y^2))/z^2

Answer 14

All of the parentheses are confusing me there.

Answer 15

I will give you another example then:

\[\frac{c^2y^{-2}}{c^{-3}y^{3}} = c^2c^{3}y^{-2}y^3\]

so for c variable,
2 + 3 = 5

for the y variable
-2 + (3) = 1

so we have,

\[c^{5}y^{1} = c^{5}y\]

Answer 16

*Hint*

Any variable raised to the power 0 is equal to 1
\[x^0 = 1\]

Answer 17

Okay I get it now, thanks!

Answer 18

You just need to apply the two rules i gave you and these problems become super easy

Answer 19

\[x^{-1} = \frac{1}{x^{1}}\]
and
\[x^{1} = \frac{1}{x^{-1}}\]

Remembering,
\[x^1 = x\]