Solve for t:
t^6 = 8-t^3
Subtract 8-t^3 from both sides:
t^6+t^3-8 = 0
Substitute x = t^3:
x^2+x-8 = 0
Add 8 to both sides:
x^2+x = 8
Add 1/4 to both sides:
x^2+x+1/4 = 33/4
Write the left hand side as a square:
(x+1/2)^2 = 33/4
Take the square root of both sides:
x+1/2 = sqrt(33)/2 or x+1/2 = -sqrt(33)/2
Subtract 1/2 from both sides:
x = sqrt(33)/2-1/2 or x+1/2 = -sqrt(33)/2
Substitute back for x = t^3:
t^3 = sqrt(33)/2-1/2 or x+1/2 = -sqrt(33)/2
Taking cube roots gives (sqrt(33)/2-1/2)^(1/3) times the third roots of unity:
t = -(1/2 (1-sqrt(33)))^(1/3) or t = (1/2 (sqrt(33)-1))^(1/3) or t = (-1)^(2/3) (1/2 (sqrt(33)-1))^(1/3) or x+1/2 = -sqrt(33)/2
Subtract 1/2 from both sides:
t = -(1-sqrt(33))^(1/3)/2^(1/3) or t = (sqrt(33)-1)^(1/3)/2^(1/3) or t = ((-1)^(2/3) (sqrt(33)-1)^(1/3))/2^(1/3) or x = -1/2-sqrt(33)/2
Substitute back for x = t^3:
t = -(1-sqrt(33))^(1/3)/2^(1/3) or t = (sqrt(33)-1)^(1/3)/2^(1/3) or t = ((-1)^(2/3) (sqrt(33)-1)^(1/3))/2^(1/3) or t^3 = -1/2-sqrt(33)/2
Taking cube roots gives (-1/2-sqrt(33)/2)^(1/3) times the third roots of unity:
Answer: | t = -(1-sqrt(33))^(1/3)/2^(1/3) or t = (sqrt(33)-1)^(1/3)/2^(1/3) or t = ((-1)^(2/3) (sqrt(33)-1)^(1/3))/2^(1/3) or t = (1/2 (-1-sqrt(33)))^(1/3) or t = -(1/2 (1+sqrt(33)))^(1/3) or t = -((-1)^(2/3) (1/2 (1+sqrt(33)))^(1/3))
