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

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We have to solve for x given 64x^3+1331=0

64x^3 + 1331 = 0

=> (4x)^3 + 11^3 = 0

=> (4x + 11)( (4x)^2 - 4x*11 + 11^2) = 0

4x + 11 = 0 => x1 = -11/4

=> x = -2.75

( (4x)^2 - 4x*11 + 11^2) = 0

x2 = [44 + sqrt( 44^2 - 4*11^2*16)]/32

=> x2 = 44/32 + i*2.3815

=> x2 = 11/8 + i*2.3815

=> x2 = 1.375 + i*2.3815

x3 = x2 = [44 - sqrt( 44^2 - 4*11^2*16)]/32

=> x3 = 11/8 - i*2.3815

=> x3 = 1.375 - i*2.3815

The values of x are **-2.75 , 1.375 + i*2.3815 , 1.375 - i*2.3815**

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 }.**