<PRE><B>Question 155177</B>
a,b,c are three distinct real numbers and there are real numbers x,y such that a&lt;sup&gt;3&lt;/sup&gt;+ax+y=0,b&lt;sup&gt;3&lt;/sup&gt;+bx=y=0 and c&lt;sup&gt;3&lt;/sup&gt;+cx+y=0 Show that a+b+c=0

&lt;pre&gt;&lt;font size = 4 color =&quot;indigo&quot;&gt;&lt;b&gt;
All three equations are cases of the cubic 
polynomial equation:

z&lt;sup&gt;3&lt;/sup&gt;+xz+y=0

Since a&lt;sup&gt;3&lt;/sup&gt;+ax+y=0,b&lt;sup&gt;3&lt;/sup&gt;+bx=y=0 and c&lt;sup&gt;3&lt;/sup&gt;+cx+y=0, this shows
that z=a, z=b and z=c are distinct solutions to 

z&lt;sup&gt;3&lt;/sup&gt;+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

z&lt;sup&gt;3&lt;/sup&gt;+xz+y=0

Therefore its left side must factor as

(z-a)(z-b)(z-c)=0

(z-a)(z&lt;sup&gt;2&lt;/sup&gt;-cz-bz+bc)=0

z&lt;sup&gt;3&lt;/sup&gt;-cz&lt;sup&gt;2&lt;/sup&gt;-bz&lt;sup&gt;2&lt;/sup&gt;+bcz-az&lt;sup&gt;2&lt;/sup&gt;+acz+abz-abc=0

z&lt;sup&gt;3&lt;/sup&gt;-(a+b+c)z&lt;sup&gt;3&lt;/sup&gt;+(ab+ac+bc)z-abc=0

This must be the same equation as

z&lt;sup&gt;3&lt;/sup&gt;+xz+y=0

Therefore the coefficient of z&lt;sup&gt;2&lt;/sup&gt; in

z&lt;sup&gt;3&lt;/sup&gt;-(a+b+c)z&lt;sup&gt;2&lt;/sup&gt;+(ab+ac+bc)z-abc=0

must be 0 since z&lt;sup&gt;3&lt;/sup&gt;+xz+y=0 has no
term in z&lt;sup&gt;2&lt;/sup&gt;.  Therefore

-(a+b+c)=0 or

a+b+c=0.

That&#39;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&lt;/pre&gt;&lt;/font&gt;&lt;/b&gt;
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