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

from Vieta's Formulas we have

y1+y2+y3=0

y1y2+y2y3+y1y3=b

y1y2y3=-b

also we have y^3=-b(y+1)

combining the fractions in the equation we get

[y1^3y3+y2^3y1+y3^3y2]/(y1y2y3)=-8

substituting we get

[-b(y1+1)y3-by1(y2+1)-by2(y3+1)/-b=-8

y3(y1+1)+y1(y2+1)+y2(y3+1)=-8

(y1+y2+y3)+(y1y2+y2y3+y1y3)=-8

thus

b=-8

thus

y^3-8y-8=0

y= 1+sqrt(5), 1-sqrt(5), -2 thus

y1=1+sqrt(5)

y2=1-sqrt(5)

y3=-2

Posted by Daniel on 2008-02-03 12:11:50