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
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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 }.
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