For the second solution

Actually there are 3 solutions:

The "second" solution composed of two:

[1 +i*sqrt(3)]/4

and [1 -i*sqrt(3)]/4

It's complicated to check with the plus-or-minus, so let's check with the plus,

x = [1 +i*sqrt(3)]/4

8x^3 +1 = 0

8{[1 +sqrt(-3)]/4}^3 +1 =? 0

We can use the formula (a+b)^3 = a^3 +3(a^2)(b) +3(a)(b^2) +b^3

but that is complicated too, or very long.

So we use (a+b)^3 = (a+b)(a+b)^2

{[1 +sqtrt(-3)]/4}^2

= (1/16){1 +2sqrt(-3) -3}

= (1/16){-2 +2sqrt(-3)}

= (1/8){-1 +sqrt(-3)}

Multiply that by [1 +sqrt(-3)]/4,

= (1/8){-1 +sqrt(-3)} * [1 +sqrt(-3)]/4

= (1/32){-1*1 -1*sqrt(-3) +sqrt(-3) -3}

= (1/32){-4}

= -1/8

So,

8x^3 +1 = 0

8{[1 +sqrt(-3)]/4}^3 +1 =? 0

8{-1/8} +1 =? 0

-1 +1 =? 0

0 =? 0

Yes, so, OK, [1 +i*sqrt(3)]/4 is a root.

Try the x = [1 -i*sqrt(3)]/4.

It should check too.