Simplifying Polynomials: [(2x^7y^2)/(4xy^3)]^3

John Whitaker

Junior Member
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May 9, 2006
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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
 
So is this the polynomial?

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

Also you might want to incude all 4 of the answer choices.
 
None of the you gave works.
The answer should be:
\(\displaystyle \L
\frac{{x^{18} }}{{8y^3 }}\)
 
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
 
Hello, John!

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

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

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

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

[I bet you looked at "2 cubed" and said "three 2's are six".]
 
Got it! Thanks. I was multiplying 2 by 3... when I should have been cubing 2. Thanks again.
John
 
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