^{3}

^{3})

^{2}*y)

^{2}which can be pulled out by multiplying the term in front of the radical by y. This gives us

^{2}√(3y)

^{1/2}=y

^{2/2+1/2}=y

^{3/2}

^{3/2}[1+2y]

Add assume that all variables represent positive real numbers

3y√27y+6y√27y^{3}

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3y√(27y)+6y√(27y^{3})

First we can simplify the expressions in the radicals to bring out terms.

3y√(9*3*y)+6y√(9*3*y^{2}*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 y^{2} which can be pulled out by multiplying the term in front of the radical by y. This gives us

9y√(3y)+18y^{2}√(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*y^{1/2}=y^{2/2+1/2}=y^{3/2}

Thus, another form of the answer is

9√(3)y^{3/2}[1+2y]

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