a,b,c are three distinct real numbers and there are real numbers x,y such that a3+ax+y=0,b3+bx=y=0 and c3+cx+y=0 Show that a+b+c=0
All three equations are cases of the cubic
polynomial equation:
z3+xz+y=0
Since a3+ax+y=0,b3+bx=y=0 and c3+cx+y=0, this shows
that z=a, z=b and z=c are distinct solutions to
z3+xz+y=0
and since that equation is of degree 3 it can have no
more than 3 solutions. Thus z=a, z=b, and z=c are the
three real solutions to
z3+xz+y=0
Therefore its left side must factor as
(z-a)(z-b)(z-c)=0
(z-a)(z2-cz-bz+bc)=0
z3-cz2-bz2+bcz-az2+acz+abz-abc=0
z3-(a+b+c)z3+(ab+ac+bc)z-abc=0
This must be the same equation as
z3+xz+y=0
Therefore the coefficient of z2 in
z3-(a+b+c)z2+(ab+ac+bc)z-abc=0
must be 0 since z3+xz+y=0 has no
term in z2. Therefore
-(a+b+c)=0 or
a+b+c=0.
That's all you were asked to show.
However by equating the other terms,
you can also find x and y:
x=ab+ac+bc and y=-abc.
Edwin