SOLUTION: <pre> How would I simplify this cubed problem? 3 ___________ ____ \)-432x^17y^12 </pre>

Algebra ->  Radicals -> SOLUTION: <pre> How would I simplify this cubed problem? 3 ___________ ____ \)-432x^17y^12 </pre>      Log On


   



Question 307213:
How would I simplify this cubed problem? 3 ________________
                                         \)-432x^17y^12


Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!

That's a clever way you came up with to make a cube
root radical.  But you can put this

root(3,-432x^17y^12)

between triple brackets (three of these "{" in a row on the 
left and three of these "}" in a row on the right and you get
this posted:

root%283%2C-432x%5E17y%5E12%29

Any odd root of a negative number is negative so
we first of all take the negative sign out.  (You
can only do this with odd roots, not even roots).
So first we take out the negative sign:

-root%283%2C432x%5E17y%5E12%29

Next we break 432 into prime factors:



-root%283%2C2%5E4%2A3%5E3x%5E17y%5E12%29

Divide the index 3 into each exponent:

Divide index 3 into 4, the exponent of 2

  1
3)4
  3
  1

The quotient is 1 and the remainder is 1

The quotient tells us what power of 2 comes out of the
radical and the remainder tells us what power of 2 remains
under the radical.  So 2%5E1 comes out in front of the
radical and 2%5E1 remains under the radical:

-2%5E1%2Aroot%283%2C2%5E1%2A3%5E3x%5E17y%5E12%29

and of course we can erase the 1 exponents:

-2%2Aroot%283%2C2%2A3%5E3x%5E17y%5E12%29

Divide index 3 into 3, the exponent of 3

  1
3)3
  3
  0 

The quotient is 1 and the remainder is 0

The quotient tells us what power of 3 comes out of the
radical and the remainder tells us what power of x remains
under the radical.  So 3%5E1 comes out in front of the
radical and no power of 3 remains under the radical, or
you can say 3%5E0 which is just 1 remains under the 
radical, but we don't even have to write it:

-2%2A3%2Aroot%283%2C2x%5E17y%5E12%29

Now we can multiply the -2 and the 3 in front of the radical
and get:

-6%2Aroot%283%2C2x%5E17y%5E12%29

Divide index 3 into 17, the exponent of x

   5
3)17
  15
   2 

The quotient is 5 and the remainder is 2

The quotient tells us what power of x comes out of the
radical and the remainder tells us what power of x remains
under the radical.  So x%5E5 comes out in front of the
radical and x%5E2 remains under the radical:

-6x%5E5%2Aroot%283%2C2x%5E2y%5E12%29

Divide index 3 into 12, the exponent of y

   4
3)12
  12
   0 

The quotient is 4 and the remainder is 0

The quotient tells us what power of y comes out of the
radical and the remainder tells us what power of y remains
under the radical.  So y%5E4 comes out in front of the
radical and no power of y remains under the radical, or
you can say x%5E0 which is just 1 remains under the 
radical, but we don't even have to write it:

-6x%5E5y%5E4%2Aroot%283%2C2x%5E2%29

Edwin