The cubic equation y³ + by + b = 0 has three roots y₁, y₂ and y₃ with y₁ ≥ y₂ ≥ y₃, where b is real and non zero, such that:

y₁²/ y₂ + y₂²/ y₃ + y₃²/ y₁= -8

Determine y₁, y₂ and y₃

y1³/y1y2 + y2³/y2y3 + y3³/y1y3 = -8
substitute y³=-b(y+1), we get
(y1+1)/y1y2 + (y2+1)/y2y3 + (y3+1)/y1y3 = 8/b
1/y1+1/y2+1/y3+1/y1y2+1/y2y3+1/y3y1 = 8/b
1/y1,1/y2,1/y3 are roots of by³+by²+1=0
1/y1+1/y2+1/y3 = -1
1/y1y2+1/y2y3+1/y3y1=0
So, 8/b=-1 => b=-8
y³-8y-8=0 is the cubic equation
=> (y+2)(y²-2y-4)=0
=> y=-2,1+√5 and 1-√5 are the solutions.

Edited on February 3, 2008, 6:19 am

Posted by Praneeth on 2008-02-02 16:16:28