What is x if 64x^3+1331=0?

To solve the binomial equation, we'll apply the formula of
the sum of cubes:

a^3 + b^3 = (a+b)(a^2 - ab +
b^2)

a^3 = 64x^3

a =
4x

b^3 = 11^3

b =
11

64x^3+1331 = (4x+11)(16x^2 - 44x +
121)

If 64x^3+1331 = 0, then (4x+11)(16x^2 - 44x + 121) =
0

If a product is zero, then each factor could be
zero.

4x+11 = 0

We'll
subtract 11 both sides:

4x =
-11

x1 = -11/4

16x^2 - 44x +
121 = 0

We'll apply the quadratic
formula:

x2 = [44 +
sqrt(1936-7744)]/2

x2 =
(44+44isqrt3)/32

We'll factorize by 44 the
numerator:

x2 =
44(1+isqrt3)/32

x2 =
11(1+isqrt3)/8

x3 =
11(1-isqrt3)/8

The roots of the equation are:
{-11/4 , 11(1-isqrt3)/8, 11(1+isqrt3)/8 }.