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John Whitaker
07-07-2006, 05:19 PM
The problem is: (2x^7y^2 / 4xy^3)^3
I get an answer: x^18 / 6y^3... but that answer is not one of the 4 to be considered. The closest one is: x^18 / 8y^3. The others seem too far out. Thanks. John

jonboy
07-07-2006, 05:28 PM
So is this the polynomial?

\L\bold \left( {\frac{{2x^{7y^2 } }}{{4xy^3 }}} \right)^3

Also you might want to incude all 4 of the answer choices.

John Whitaker
07-07-2006, 05:35 PM
Yes. Thank you.
1. x^18y^3 / 2
2. x^6 / 8y^3
3. 8x^18y^6

John Whitaker
07-07-2006, 05:38 PM
Staring at the 3 answers I just entered, #1 looks promising. John

pka
07-07-2006, 05:48 PM
None of the you gave works.
The answer should be:
\L
\frac{{x^{18} }}{{8y^3 }}

John Whitaker
07-07-2006, 05:55 PM
pka, thank you. Can you tell me how the "8" comes into it? John

John Whitaker
07-07-2006, 06:20 PM
I get:
2*x*x*x*x*x*x*x*y*y / 2*2*x*y*y*y... and after cancelling, I get:
(x^6) / 2y)^3
3*x^6=x^18
3*2y=6y^3
How do you get 8y^3
John

soroban
07-07-2006, 06:24 PM
Hello, John!


\L\left(\frac{2x^7y^2}{4xy^3\right)^3

I get an answer: \L\frac{x^{18}}{6y^3} . . . no
but that answer is not one of the 4 to be considered.
The closest one is: \L\frac{x^{18}}{8y^3} . . . yes
Inside the parenthese, reduce the fraction:

\;\;\L\left(\frac{2x^7y^2}{4xy^3}\right)^3\;=\; \left(\frac{x^6}{2y}\right)^3

Then "cube everything": \L\:\frac{(x^6)^3}{2^3y^3} \;=\;\frac{x^{18}}{8y^3}

[I bet you looked at "2 cubed" and said "three 2's are six".]

John Whitaker
07-07-2006, 06:30 PM
Got it! Thanks. I was multiplying 2 by 3... when I should have been cubing 2. Thanks again.
John