How do you simplify #6root4 9 +2 root3 9 +2 root4 9 - 3 root3 9#?

1 Answer
Mar 29, 2016

Answer:

# sqrt3(+-6+-2)-9^(1/3)# = #sqrt3(+-6+-2)-2.08)#, nearly.
The four values are #11.78, 4.85, -9.01 and -15.94#, nearly.

Explanation:

#9^(1/4)=(9^(1/2))^(1/2)=(+3)^(1/2)=+-sqrt3#.

The first and third tems add up to #sqrt3(+-6+-2)#.

The second and fourth add up to #-9^(1/3#..

So, the expression simplifies to #sqrt3(+-6+-2)-9^(1/3)#.
There is only one real value 2.08, nearly, for #9^(1/3).

In general, #a^(1/n)# has n values. The complex values occur in conjugate pairs. For example, the values #1^(1/3)#are #1, +-i#.,