# Let : $P(x)=x^{3}+ax^{2}+bx+c$ where $(a,b,c)\in Z^3$

Question :

If :

$$P(x)=x^3+ax^2+bx+c$$ where $$(a,b,c)\in Z^3$$

And $$m,n,k$$ root of $$P(x)$$

such that : $$m.n=k$$

Then show that : $$2P(-1)$$ multiple of

$$P(1)+P(-1)-2[1+P(0)]$$

My try :

We known that :

$$m+n+k=-a$$

$$mn+mk+nk=b$$

$$m.n.k=-c$$

Since $$m.n=k$$

So : $$k^2=-c$$

But how I complete this work

Please give me ideas or hints

First let's simplify the things that are required in the proof.

$$P(1) + P(-1) - 2[1+P(0)] = (1+a+b+c)+(-1+a-b+c) -2(1+c)$$

$$=2a-2$$

And,

$$2P(-1) = 2(-1+a-b+c)$$

Now we have to show that $$2a-2$$ divides $$2(-1+a-b+c)$$ which is equivalent to saying that $$a-1$$ divides $$b-c$$.(Why?)

Now writing out the expressions for $$a,b,c$$.

$$a = -(m+n+k)$$ $$b=mn+nk+mk$$ $$c = -mnk$$

Therefore,

$$b-c= mn+nk+mk+mnk$$ $$=k+nk+mk+k^2$$

$$=k(1+m+n+k)=k(1-a)$$

Which is what we wanted!

• Very good ideas , brilliant Sir – Kînan Jœd May 18 at 0:13
• Sure, no problem. Plz mark the answer as accepted if you have no further queries. – Vizag May 18 at 17:46