SOLUTION: cube root of 27w to the 12th power over 100 {{{cube( 27w^12/100 ) }}} Explanation: I am a volunteer math tutor. My student disagrees with the answer I came up with, so I

Algebra ->  Square-cubic-other-roots -> SOLUTION: cube root of 27w to the 12th power over 100 {{{cube( 27w^12/100 ) }}} Explanation: I am a volunteer math tutor. My student disagrees with the answer I came up with, so I       Log On


   

Question 243853: cube root of 27w to the 12th power over 100
cube%28+27w%5E12%2F100+%29+
Explanation:
I am a volunteer math tutor. My student disagrees with the answer I came up with, so I am writing to you for any help in justifying her answer or my answer as "more correct". Here's the problem: Show your work in solving this cube root problem.
The heart of the question is my student's answer vs. my answer. Which one is more correct?
Student:
cube( (3 x 3 x 3) w^(4 x 3)/(75 + 25) )
3w4/(5 x 5 x 5) + cube( 25 )
3w4/5 + cube ( 25 ) <<< student's answer
Tutor:
cube( (3 x 3 x 3) w^(4 x 3)/(75 + 17 + 8) )
3w4/(5 x 5 x 5) + cube( 17 ) + (2 x 2 x 2)
3w4/5 + cube ( 17 ) + 2
3w4/7 + cube (17) <<< tutor's answer
Which answer will the teacher of my student consider as "correct" ?
3w4/5 + cube ( 25 ) OR 3w4/7 + cube (17)
Thank you. Rob Miller
Found 2 solutions by scott8148, MathTherapy:
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
breaking the denominator into sums is NOT a good technique for simplifying this

my two cents:

multiplying the fraction by 10/10 ___ cbrt[(10 * 27w^12) / 1000]

extracting WHOLE roots ___ (3w^4 / 10)cbrt(10)

this leaves no roots in the denominator

Answer by MathTherapy(10557) About Me  (Show Source):
You can put this solution on YOUR website!
cube root of 27w to the 12th power over 100

cube%28+27w%5E12%2F100+%29+

Explanation:

I am a volunteer math tutor.  My student disagrees with the answer I came up with, so I am writing to you for any help in
justifying her answer or my answer as "more correct".  Here's the problem:  Show your work in solving this cube root problem.

The heart of the question is my student's answer vs. my answer.  Which one is more correct?

Student:
cube( (3 x 3 x 3) w^(4 x 3)/(75 + 25) )
 3w4/(5 x 5 x 5) + cube( 25 )
 3w4/5 + cube ( 25 ) <<< student's answer

Tutor:
cube( (3 x 3 x 3) w^(4 x 3)/(75 + 17 + 8) )
 3w4/(5 x 5 x 5) + cube( 17 ) + (2 x 2 x 2)
 3w4/5 + cube ( 17 ) + 2
 3w4/7 + cube (17) <<< tutor's answer

Which answer will the teacher of my student consider as "correct" ?

 3w4/5 + cube ( 25 ) OR 3w4/7 + cube (17)

Thank you.  Rob Miller

Neither, I'm afraid to say!! One error I noticed: 5x5x5+%3C%3E+75.
Plus, the denominator cannot be split the way you two did!

root+%283%2C+%2827w%5E12%29%2F100%29 
root+%283%2C+%2810%2F10%29%28%2827w%5E12%29%2F%28100%29%29%29 ----- Multiplying RADICAND by 1: 10%2F10
root+%283%2C+10%283w%5E4%29%5E3%2F10%5E3%29 = %283w%5E4%2F10%29root+%283%2C+10%2F1%29 = highlight%28%283w%5E4%2F10%29%28root+%283%2C+10%29%29%29

OR

root+%283%2C+%2827w%5E12%29%2F100%29 = root+%283%2C+%283w%5E4%29%5E3%2F100%29 = %283w%5E4%29%2Aroot%283%2C+%281%2F100%29%29 = highlight%28%283w%5E4%29%2Aroot%283%2C+10%5E%28-+2%29%29%29%29, or highlight%28%283w%5E4%29%2Amatrix%282%2C1%2C+%22+%22%2C+%2810%5E%28-+2%2F3%29%29%29%29%29