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