2012/012) simplify ³√[(√980/27) + 6] - ³√[(√980/27) - 6]?

let x = ³√(√980/27) + 6

and y = ³√- ([(√980/27) - 6)


x^3 + y^3 = 12
xy = - ³√(980/27-36) = - ³√(980/27-36) = - ³√8/27 = - 2/3
we need to find x + y

(x^3+y^3) = (x+y)^3 - 3xy(x+y)
or 12 = (x+y)^3 + 2(x+y)
or (x+y)^3 + 2(x+y) - 12 = 0

if x + y = t

t^3 + 2t - 12 = 0

t = 2 is a real root and and other roots are complex

hence we have ³√[(√980/27) + 6] - ³√[(√980/27) - 6] = 2