SOLUTION: cubed root of 10x^2y^4 * cubed root of 4x^2y^2

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Question 35292This question is from textbook Prentice Hall Mathematics ALgebra 2
: cubed root of 10x^2y^4 * cubed root of 4x^2y^2 This question is from textbook Prentice Hall Mathematics ALgebra 2

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
root%283%2C10x%5E2y%5E4%29+%2A+root%283%2C4x%5E2y%5E2%29+

Multiply the quantities inside the cuberoots:
root%283%2C10x%5E2y%5E4%2A4x%5E2y%5E2%29+
root%283%2C40x%5E4y%5E6%29

Sort out the perfect cubes and place them in the first cuberoot sign, and the leftover factors go in the second cuberoot. Remember that the perfect cube that divides into 40 is 8. Also remember that exponents must divide by 3.
root%283%2C8x%5E3y%5E6%29+%2A+root%283%2C5x%29+
+2xy%5E2+%2Aroot%283%2C5x%29

R^2 at SCC