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let the common root be p ... q,r,s,t be other roots
f(x) = a(x-p)(x-q)(x-r)
g(x) = c(x-p)(x-s)(x-t)
equating to zero, you get
ax^3 - a(p+q+r)x^2 + (pq+qr+rp)ax + pqra = ax^3+bx+c
do the same for other, eliminate q, and r comparing coefficients
and from other eliminate s and t,
from both equations you get interms of p, equate 'p' from both and you have your condition. have to see if this get's complicated or not.
the condition depends on your initial parameters like 'a', 'b', 'c'.
so you have to reduce it to the form in a,b,c which it must satisfy in order to have a common root.
you may use http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method to solve for cubic. just find one root and keep squaring or cubic unitl you get rid of surds.
that's how i old you
ax^3 - a(p+q+r)x^2 + (pq+qr+rp)ax + pqra = ax^3+bx+c
p+q+r = 0
pq +qr +rp = b/a
pqra = c
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from there equations eliminate q, and r ... you will get an equation in terms of p,a,b,c
similarly from here
c(x-p)(x-s)(x-t) = cx^3 +bx^2 +a
eliminate s and t, you should get equation in p,a,b,d
isolate p from both sides .. .this one and another one ... and equal them. you should get the required condition on 'a','b','c'
let the common root be y then,
ay^3+by+c=0 ....i
cy^3+by^2+a=0 ...ii
subtract ii from i,
(a-c)y^3 + b(y-y^2) + (c-a)=0
(a-c)(y^3 -1) + by(1-y)=0
(a-c)(y-1)(y^2+y+1)-by(y-1)=0
(a-c) (y^2+y+1) -by=0
(a-c)y^2 + (a-c-b)y +(a-c)=0
Now, y exists only if (a-c-b)^2 >= 4*(a-c)*(a-c)
They cannot only have only one common root imaginery...... if one common root is imaginery then both imaginery roots must be common.
But not sure how this would work