Solve for x in complex numbers set: x^6 - 7x^3 - 8 = 0

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Let x^3 be t:

We'll re-write the equation:

t^2 - 7t - 8 = 0

We'll apply quadratic formula:

t1 = [7+sqrt(49 + 32)]/2

t1 = (7+9)/2

t1 = 8

t2 = -1

x^3 = 8 => x^3 - 8 = 0

The difference of cubes returns the product:

x^3 - 8 =(x-2)(x^2 + 2x + 4)

If x^3 - 8 = 0 => (x-2)(x^2 + 2x + 4) = 0

x - 2 = 0 => x = 2

x^2 + 2x + 4 = 0

We'll apply quadratic formula:

x1 = [-2+sqrt(4-16)]/2

x1 = (-2 + 2i*sqrt3)/2

x1 = -1+i*sqrt3

x2 = -1-i*sqrt3

x^3 = -1 => x^3 + 1 = 0

The difference of cubes returns the product:

x^3 + 1 = (x+1)(x^2 - x + 1)

We'll cancel each factor:

x + 1 = 0 => x = -1

x^2 - x + 1 = 0

We'll apply quadratic formula:

x1 = [1+sqrt(1-4)]/2

x1 = (1+isqrt3)/2

x2 = (1-isqrt3)/2

The complex roots of equation are: {-1 ; 2 ; -1+i*sqrt3 ; -1-i*sqrt3 ; (1+isqrt3)/2 ; (1-isqrt3)/2}.

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