Question

Officially to @zepdrix

Answer 1

Why me? XD lol

Answer 2

Well, because

you're smart
and healthy

Answer 3

anyways my problem is kinda longer this time

Answer 4

\[(\frac{ -8x^6 }{ y^6-3} )^\frac{ 2 }{ 3 }\]

Answer 5

well and so my answer is

Answer 6

\(\frac{4 (-1)^{2/3} x^4}{y^2}\) tf it doesnt works Smh.

Answer 7

o.m.g. nvm it does xP

Answer 8

Anyways so this is my answer.. ;3

Answer 9

@zepdrix @zepdrix @zepdrix

Answer 10

This is the expression?\[\large\rm \left(\frac{-8x^6}{y^6-3}\right)^{\frac23}\]Hmm we can't do very much with the denominator.. this is a strange problem D:

Answer 11

Was the -3 supposed to be in the exponent?

Answer 12

Fml.
i was writing \[(\frac{ -8x^6 }{ y^6 }-3)^2\]

Answer 13

Oh oh I see.

Answer 14

So this? wrong/right

Answer 15

\[\large\rm \left(\frac{-8x^6}{y^6}-3\right)^2\]So expanding out the binomial,\[\large\rm \left(\frac{-8x^6}{y^6}\right)^2+2(-3)\frac{-8x^6}{y^6}+(-3)^2\]

Answer 16

._.

Answer 17

How did you end up with only y^2 in the bottom?
Hmm

Answer 18

i got 12 at first

Answer 19

Squaring everything in the first term,\[\large\rm \frac{16x^{12}}{y^{12}}+2(-3)\frac{-8x^6}{y^6}+(-3)^2\]

Answer 20

Simplifying the second term,\[\large\rm \frac{64x^{12}}{y^{12}}+\frac{48x^6}{y^6}+(-3)^2\]and the third term,\[\large\rm \frac{64x^{12}}{y^{12}}+\frac{48x^6}{y^6}+9\]

8 squared is not 16
mistake on my last post :)

Answer 21

Oh okay._.
so i hae to solve this?

Answer 22

*solve

Answer 23

Yes, you could simplify it further by getting a common denominator I suppose.

Answer 24

KK this might take me forever.. but oh well

Answer 25

Just the third term?

Answer 26

Hmm how bout try to match the first and second denominators to start.
First has a y^12,
second has a y^6

so how do we fix that? :d

Answer 27

So like this?
\[\frac{ y ^{12} }{ y^6 }\]
which would be d

Answer 28

i mean *2

Answer 29

What? 0_o hmm

Answer 30

huh x.x

Answer 31

That's for the second term?
We need to multiply top and bottom by y^6 to get the y^12 that we need, yes?\[\large\rm \frac{64x^{12}}{y^{12}}+\frac{48x^6}{y^6}\color{royalblue}{\frac{y^6}{y^6}}+9\]

Answer 32

those blue thingies?

Answer 33

Oh oh, ya know what would be a lot easier?
Why don't do the trouble of getting a common denominator `before expanding the square`.

Answer 34

1?

Answer 35

\[\large\rm \left(\frac{-8x^6}{y^6}-3\right)^2\]The 3 needs a y^6 top and bottom, yes?\[\large\rm \left(\frac{-8x^6}{y^6}-\frac{3y^6}{y^6}\right)^2\]And then we can rewrite it as a single fraction,\[\large\rm \left(\frac{-8x^6-3y^6}{y^6}\right)^2\]Do those steps make sense? :o

Answer 36

O yes, now they do.

Answer 37

\[\large\rm \left(\frac{-8x^6-3y^6}{y^6}\right)^2\quad=\quad \frac{\qquad\qquad\qquad\qquad\qquad}{y^{12}}\]So when we square the denominator, we should get y^12.
What do we get up top?

Answer 38

K so i solve the top right now?

Answer 39

Yes, square the top :U

Answer 40

Oh mfg, i cant simplify it

Answer 41

You're just dealing with this,\[\large\rm (-8x^6-3y^6)^2\]Here is a little trick you can do if it's not too confusing:
Factor a negative out of each term,\[\large\rm (-(8x^6+3y^6))^2\]And then square the negative,\[\large\rm (8x^6+3y^6)^2\]Now you don't have to worry about all of those extra minus signs.

Answer 42

"Well, because

you're smart
and healthy"

This deserved more love.