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How do you do this algebra equation?

Answer= x/1728(y^12)

1.Multiply the -3 to all the powers int top > (6^-3)(x^-3)(y^-9)
2. subtract the powers from top to bottom
3. make all powers positive by moving negative powers to bottom
4. 8*(6^3)=1728
 
7window said:
Answer= x/1728(y^12)

1.Multiply the -3 to all the powers int top > (6^-3)(x^-3)(y^-9)
2. subtract the powers from top to bottom
3. make all powers positive by moving negative powers to bottom
4. 8*(6^3)=1728
Click to expand...

That isn't an answer, you simply rewrote the 'equation'.
 
Being put in with a proper format would be a good start.
The next important step is actually setting it equal to something so it really is an equation.
 
(6xy^3)^-3

First, multiply-through the 3.
(18xy^9)^-

Then the -.
(-18xy^9)

Then the ^.
(-18xy^²9)

Then the parentheses cancel each other out.
-18xy^²9

Expand the ^, and rearrange.
^-8xy^19.

Rotate the 8, and move the - down a bit.
^_∞xy^19

Set the x & y aside for a bit, and split the ∞ apart.
Rearrange, and you arrive at your solution:

I dunno LOL
^o_o^







Edit: Ah, now I see the second part. I completely missed the:
_______
(8x^-4y^3)
, having mistook it for a signature, and thus not related to the post. Derp.
I still prefer my solution though. And yes, 'tis not an equation if there's no equals sign. 😛


.
 
Last edited:
MSP2331a03c1h8b3dhf254000050fb563649da2897
 
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