
William R. answered 10/04/13
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MIT Grad in Physics for Math, Science and More!
3y√(27y)+6y√(27y3)
First we can simplify the expressions in the radicals to bring out terms.
3y√(9*3*y)+6y√(9*3*y2*y)
The 27 in both radicals have a factor of 9, which is the square of 3. We can bring out this factor by multiplying the terms in front of the radicals by 3. The second term has a factor of y2 which can be pulled out by multiplying the term in front of the radical by y. This gives us
9y√(3y)+18y2√(3y)
Now we can factor out the common term 9y√(3y) from both parts of the expression. This gives our answer
9y√(3y)[1+2y]
We can further simplify this by combining the y and √y in front of the brackets into a single term. y*y1/2=y2/2+1/2=y3/2
Thus, another form of the answer is
9√(3)y3/2[1+2y]