SOLUTION: I am preparing my self for the upcoming next level algebra semester in the fall. I have a practice test with some problems I will encounter during the semester. As a reference I am
Algebra ->
Radicals
-> SOLUTION: I am preparing my self for the upcoming next level algebra semester in the fall. I have a practice test with some problems I will encounter during the semester. As a reference I am
Log On
Question 1084311: I am preparing my self for the upcoming next level algebra semester in the fall. I have a practice test with some problems I will encounter during the semester. As a reference I am using Blitzer (great book) along with the solution manual. Right of the rip I was hit with this problem, which I can not find in the book. Please tell the name of this problem so I can look it up in the book and find other similar problems to practice and show me the steps involved to solve the problem and I will figure it out. its two square roots multiplied against each other. ∛(□((x^0 y^4)/(3z^3 ))) . ∛(81x^9 y^(-10) z^6 )Simplify completely.
You can put this solution on YOUR website! The cube root of (y^4/z^3) (x^0=1) is (y/z)*y^(1/3)
The cube root of the second is 4x^3*z^2/y^3*y^(1/3)
The y^1/3 cancel and the answer is 4x^3*z/y^2
taking the cube root raises the exponent present to the (1/3) power.
This is multiplying square roots or in this case cube roots.
If they have the same index, they can be multiplied together directly. If the index is different, it can be converted to a fractional exponent. sqrt (x)* cube rt (x)=x^(1/2)*x^(1/3)=x^(5/6)
