SOLUTION: 4^√243x^8y^10 my answer was (6x^3y^5)^4√3x^2 but the answer in my math book is (3x^2y^2)4^√3y^2 what am i doing work here? please help me. thank you.

Algebra ->  Radicals -> SOLUTION: 4^√243x^8y^10 my answer was (6x^3y^5)^4√3x^2 but the answer in my math book is (3x^2y^2)4^√3y^2 what am i doing work here? please help me. thank you.      Log On


   

Question 383778: 4^√243x^8y^10 my answer was (6x^3y^5)^4√3x^2 but the answer in my math book is (3x^2y^2)4^√3y^2 what am i doing work here? please help me. thank you.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
root%284%2C+243x%5E8y%5E10%29
I'm not sure I can figure out how you got the answer you did. So I will not be able to explain what went wrong. But your Math book is correct.

Your expression is a 4th root. And to simplify a 4th root you look for factors of the radicand (the expression within the radical) that are 4th powers. (Other factors are of no real interest.)

For 243 there may be many ways to factor it but one way, 81*3, involves a factor that is a 4th power. (81+=+3%5E4) For the variables we factor each one into as many powers of 4 as we can. Putting all these factors together we get:
root%284%2C+3%5E4%2A3%2Ax%5E4%2Ax%5E4%2Ay%5E4%2Ay%5E4%2Ay%5E2%29
Take a moment to make sure you see how your original radicand and the factored one you see above are equal to each other.

Next, I like to use the Commutative Property of Multiplication to reorder the factors so that all the 4th powers are in front:
root%284%2C+3%5E4%2Ax%5E4%2Ax%5E4%2Ay%5E4%2Ay%5E4%2A3%2Ay%5E2%29
Next we use a property of radicals, root%28a%2C+p%2Aq%29+=+root%28a%2C+p%29%2Aroot%28a%2C+q%29 to separate all the 4th power factors into their own radicals. (The factors that are not 4th powers (at the end) just stay all in the same radical.):

Now all the 4th roots of the 4th powers simplify:
3%2Ax%2Ax%2Ay%2Ay%2Aroot%284%2C+3y%5E2%29
which simplifies to:
3%2Ax%5E2%2Ay%5E2%2Aroot%284%2C+3y%5E2%29
which is the answer in your book.