Problems of this nature are intended to help you see that the use of solution techniques for quadratic equations can be extended to polynomials other than simple quadratics. Here, we can let

y = x^3

and we find that this becomes a quadratic in y

y^2 - 3y - 4 = 0

(y+1)(y-4) = 0 (factor the above equation)

Performing the reverse substitution, we find that further factoring can be done.

(x^3+1)(x^3-4) = 0

(x+1)(x^2 - x + 1)(x^3 - 4) = 0 (factor the first as the sum of two cubes)

The middle factor, a quadratic, has discriminant (-1)^2 - 4(1)(1) = -3, so it has no real roots. The remaining factors have real roots at

x = -1

x = cuberoot(4) = 4^(1/3) = 2^(2/3) ≈ 1.5874

The real roots are {-1, 2^(2/3)}.

y = x^3

and we find that this becomes a quadratic in y

y^2 - 3y - 4 = 0

(y+1)(y-4) = 0 (factor the above equation)

Performing the reverse substitution, we find that further factoring can be done.

(x^3+1)(x^3-4) = 0

(x+1)(x^2 - x + 1)(x^3 - 4) = 0 (factor the first as the sum of two cubes)

The middle factor, a quadratic, has discriminant (-1)^2 - 4(1)(1) = -3, so it has no real roots. The remaining factors have real roots at

x = -1

x = cuberoot(4) = 4^(1/3) = 2^(2/3) ≈ 1.5874

The real roots are {-1, 2^(2/3)}.