# SOLUTION: Solve this: (9x^6*y^8)^(1/6) And I have this possible answers a. xy(9y8)^(1/6) b. xy(3y)^(1/3) c. 3xy d. x(9y^2)^(1/6)

Algebra ->  -> SOLUTION: Solve this: (9x^6*y^8)^(1/6) And I have this possible answers a. xy(9y8)^(1/6) b. xy(3y)^(1/3) c. 3xy d. x(9y^2)^(1/6)      Log On

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 Click here to see ALL problems on Square-cubic-other-roots Question 40419: Solve this: (9x^6*y^8)^(1/6) And I have this possible answers a. xy(9y8)^(1/6) b. xy(3y)^(1/3) c. 3xy d. x(9y^2)^(1/6)Answer by AnlytcPhil(1321)   (Show Source): You can put this solution on YOUR website!```Solve this: (9x6y8)1/6 And I have this possible answers a. xy(9y8)1/6 b. xy(3y)1/3 <------- that's the answer! c. 3xy d. x(9y2)1/6 ======================================================== (9x6y8)1/6 Write the 9 as 32 (32x6y8)1/6 Remove the parentheses by multiplying each inner exponent by the outer exponent 1/6 3(2)(1/6)x(6)(1/6)y(8)(1/6) 32/6x6/6y8/6 Reduce the fraction exponents 31/3x1y4/3 Change improper fraction exponent 4/3 to 1 1/3 but write it as 1+1/3 31/3x1y1+1/3 Now since we add exponents of like bases to multiply, we can use that fact in reverse, and when we see an exponent consisting of an addition, such as in y1+1/3 we can write that as the multiplication y1y1/3. 31/3x1y1y1/3 Now we put the two factors which have the same exponent 1/3 together 31/3y1/3x1y1 Now since to remove parentheses we multiply the inner exponents by the outer exponents, we can use that fact in reverse to say that when two factors have the same exponents, we can put the product of the bases inside parentheses and use the common exponent outside the parentheses, so 31/3y1/3 can be written as (3y)1/3 So the above becomes (3y)1/3x1y1 We can erase the 1 exponents and get (3y)1/3xy which is the same as xy(3y)1/3 which is answer b. Edwin AnlytcPhil@aol.com```