Find all real numbers a, b, c such that x3 + ax2 + bx + c has three real roots α,β,γ (not necessarily all distinct) and the equation x3 + a3x2 + b3x + c3 has roots α3, β3, γ3.
Answer
(a,b,c) = (h,k,hk), where k ≤ 0
Solution
We have -a = α + β + γ, b = αβ + βγ + γα, -c = αβγ. We need -a3 = α3 + β3 + γ3. Since -a3 = (α + β + γ)3 = α3 + β3 + γ3 + 3(αβ2 + ... ) + 6αβγ = -a3 - 3ab + 3c, we require c = ab.
Now if c = ab, the cubic factorises as (x+a)(x2+b), so we get three real roots iff b ≤ 0. But if b ≤ 0 and c = ab, then the roots are -a, ±√(-b). So the equation with roots α3, β3, γ3 is (x + a3)(x2 + b3) = x3 + a3x2 + b3x + a3b3. Thus c = ab, b ≤ 0 is both a necessary and a sufficient condition.
© John Scholes
jscholes@kalva.demon.co.uk
6 March 2004
Last corrected/updated 6 Mar 04