To solve the equation 10x^(1/6) - 3x^(1/3) - 3 = 0, we can first let y = x^(1/6). This means y^2 = x^(1/3).

With this substitution, the equation becomes:
10y - 3y^2 - 3 = 0.

Now we can rearrange the equation:
3y^2 - 10y + 3 = 0.

We can solve this quadratic equation by factoring, using the quadratic formula, or completing the square. Let's use the quadratic formula:
y = [-(-10) ± sqrt((-10)^2 - 433)] / 2*3
y = [10 ± sqrt(100 - 36)] / 6
y = [10 ± sqrt(64)] / 6
y = (10 ± 8) / 6

The solutions for y are:
y1 = (10 + 8) / 6 = 3
y2 = (10 - 8) / 6 = 1/3

Since y = x^(1/6), we can now solve for x using these values:
For y1 = 3:
x^(1/6) = 3
x = 3^6
x = 729

For y2 = 1/3:
x^(1/6) = 1/3
x = (1/3)^6
x = 1/729

Therefore, the solutions to the equation 10x^(1/6) - 3x^(1/3) - 3 = 0 are x = 729 and x = 1/729.