First note that tan is an odd function: tan(-x) = -tan(x).  Due to the 6th powers, this will not change the value of the expression.  Apply this to the 2pi/9 argument.  Then 

tan^6(pi/9) + tan^6(2pi/9) + tan^6(4pi/9) 

= tan^6(pi/9) + tan^6(-2pi/9) + tan^6(4pi/9)

Start with the identity: 

tan(3t) = [3*tan(t) - tan^3(t)] / [1 - 3*tan^2(t)]

Substituting any of t=pi/9, t=-2pi/9, or t=4pi/9 makes tan(3t) = sqrt(3).

Rearrange the identity with tan(3t) = sqrt(3) and tan(t) = x to yield the polynomial:

x^3 - (3*sqrt(3))x^2 - 3x + sqrt(3) = 0

tan(pi/9), tan(-2pi/9), and tan(4pi/9) are the three roots of this polynomial.

Now, the Newton-Girard Formulas can be used to find the sum of squares and the sum of cubes of the three roots.

Let P(1), P(2), and P(3) be the symmetric polynomials.  Then P(1) = 3*sqrt(3), P(2) = -3, and P(3) = -sqrt(3)

Let S(1)-S(6) be the sums of the first through sixth powers of the roots.  Then:

S(1) - P(1) = 0

S(2) - S(1)*P(1) + 2*P(2) = 0

S(3) - S(2)*P(1) + S(1)*P(2) - 3*P(3) = 0

From this 

S(1) = 3*sqrt(3)

S(2) = (3*sqrt(3))*(3*sqrt(3)) - 2*(-3) = 33

S(3) = (33)*(3*sqrt(3)) - (3*sqrt(3))*(-3) + 3*(-sqrt(3)) = 105*sqrt(3)

The polynomial can be rearranged into 

x^(n+3) = (3*sqrt(3))x^(n+2) + 3x^(n+1) - sqrt(3)x^n.  

This implies the recursion 

S(n+3) = (3*sqrt(3))S(n+2) + 3S(n+1) - sqrt(3)S(n).

Then S(4) = (3*sqrt(3))*(105*sqrt(3)) + 3*(33) - sqrt(3)*(3*sqrt(3)) = 1035

And S(5) = (3*sqrt(3))*1035 + 3*(105*sqrt(3)) - sqrt(3)*(33) = 3387*sqrt(3)

And S(6) = (3*sqrt(3))*(3387*sqrt(3)) + 3*(1035) - sqrt(3)*(105*sqrt(3)) = 33273

The final answer is tan^6(pi/9) + tan^6(2pi/9) + tan^6(4pi/9)  = 33273