The cubic equation x³ + ax+ a = 0 has three roots x₁, x₂ and x₃ with x₁ ≤ x₂ ≤ x₃, where a is real and non zero, such that:

x₁²/ x₂ + x₂²/ x₃ + x₃²/ x₁ = 8

Determine x₁, x₂ and x₃

Lets say the three roots are b, c, and d (just to avoid subscripts).

Then (x-b)(x-c)(x-d) = 0  Multiplying and comparing to the 1st equation gives us:    b+c+d = 0    bc+bd+cd = a    bcd = -a      The 2nd equation becomes b^2/c + c^2/d + d^2/b = 8    Multiplying by bcd gives b^3*d + c^3*b + d^3*c = 8bcd  but we can substitute b^3 = -ab -a                        c^3 = -ac -a                        d^3 = -ad -a  and after rearranging terms,    -a(bc + bd + cd) -a(b+c+d) = 8bcd      and (using the identities above) this becomes      -a(a) -a(0) = -8a  or   a^2 = 8a    This has two solutions, a = 0 and a = 8  But a = 0 implies b = c = d = 0, which doesn't work    so, a = 8    
And, if I haven't made mistakes, then all you need 
to do is solve
   x^3 + 8x + 8 = 0
which is a known formula. 

Edited on April 25, 2014, 4:33 pm

Posted by Steve Herman on 2012-12-23 15:52:11