SOLUTION: I have the answer in the back of the book. Still having trouble getting there.
4^2/3 * 6^2/3 * 9^2/3
Have tried squaring 4 and then taking the cubed root but don't get good n
Question 51682This question is from textbook
: I have the answer in the back of the book. Still having trouble getting there.
4^2/3 * 6^2/3 * 9^2/3
Have tried squaring 4 and then taking the cubed root but don't get good numbers.
The book says answer is 36????? This question is from textbook
Found 3 solutions by mathchemprofessor, gsmani_iyer, MathTherapy:Answer by mathchemprofessor(65) (Show Source): You can put this solution on YOUR website! 4^2/3 * 6^2/3 * 9^2/3=(4^2)^(1/3)*(6^2)^(1/3)*(9^2)^(1/3)
=(16)^(1/3)*(36)^(1/3)*(81)^(1/3)
-(16*36*81)^(1/3)=(46656)^(1/3)=(36^3)^(1/3)=36
Answer by gsmani_iyer(201) (Show Source): You can put this solution on YOUR website!
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= 3rd
= 3rd = 3rd = 3rd
= 3 * 3 * 4 = 36. Answer. I hope this must have solved your trouble in getting the answer.
Note: Read 3rd as 3rd root in the above solution.
I have the answer in the back of the book. Still having trouble getting there.
4^2/3 * 6^2/3 * 9^2/3
Have tried squaring 4 and then taking the cubed root but don't get good numbers.
The book says answer is 36?????
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