To solve the first equation x^6 - x^2 = 0, we can factor out x^2:

x^2(x^4 - 1) = 0

Now, we can factor x^4 - 1 as the difference of squares:

x^2(x^2 + 1)(x^2 - 1) = 0

Now, we have three factors: x^2 = 0, x^2 + 1 = 0, and x^2 - 1 = 0.

1. x^2 = 0
x = 0

2. x^2 + 1 = 0
x^2 = -1
This equation has no real solutions, as the square of any real number is always non-negative.

3. x^2 - 1 = 0
x^2 = 1
x = ±1

Therefore, the solutions to x^6 - x^2 = 0 are x = 0, x = 1, and x = -1.

Moving on to the second equation x^2 + 4x + 1 = 0, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 1, b = 4, and c = 1.

x = (-4 ± √(4^2 - 4*1*1)) / 2*1
x = (-4 ± √(16 - 4)) / 2
x = (-4 ± √12) / 2
x = (-4 ± 2√3) / 2
x = -2 ± √3

Therefore, the solutions to x^2 + 4x + 1 = 0 are x = -2 + √3 and x = -2 - √3.

Finally, for the third equation 3x^2 + 4x - 4 = 0, we can also use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 3, b = 4, and c = -4.

x = (-4 ± √(4^2 - 4*3*(-4))) / 2*3
x = (-4 ± √(16 + 48)) / 6
x = (-4 ± √64) / 6
x = (-4 ± 8) / 6

This gives us two possible solutions:
x = (-4 + 8) / 6 = 4 / 6 = 2/3
x = (-4 - 8) / 6 = -12 / 6 = -2

Therefore, the solutions to 3x^2 + 4x - 4 = 0 are x = 2/3 and x = -2.