sorry for the trash. just disregard.

We can use Larange interpolation,so i think yes

Problem is very easy

Yes @above its easy; the required polynomial is $P(x)=x^3-(x-a)(x-b)(x-c)$.

i am sorry but read the question again it is mentioned quadratic polynomial.

please look carefully, the cubic term will be cancelled

oh crap i am less on sleep sorry someone just kill me.

<dovakin123> wrote: Yes @above its easy; the required polynomial is $P(x)=x^3-(x-a)(x-b)(x-c)$. Yes. The cubic term cancels out surely. However, how do we know the coefficient of the quadratic term is a positive integer?

- Deleted since it was wrong - I should carefully read the problem. Let me pose another problem. Prove or disprove the following; For any three different integers $a$, $b$, $c$, there is an integral coefficient quadratic polynomial which takes all three values $a^3$, $b^3$, $c^3$ at some integer points. Can we require the graph of the polynomial be concave above? At my opinion, the answer is negative and essential difficulty lies on the example like $\left\{ a,b,c \right \} = \left\{ -3,-1,0 \right\}$. From this example, we can deduce that if there is no pair of integers $(m,n,k)$ such that \[ \frac{26 k-27 m+n}{(m-k) (k-n) (m-n)} \]is a positive integer, then the answer is negative which is very likely to be true.

$a,b,c$ are positive, and $x^3-(x-a)(x-b)(x-c)=(a+b+c)x^2-(ab+ac+bc)x+abc$ has positive coefficient for $x^2$

I saw a very similar problem in 2017 AMC 12 B problem 23.[url]https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_23 [/url]