Find polynomial : $P(x)=x^3+ax^2+bx+c$

$x^3+ax^2+bx+c = (x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$

so $a=-(a+b+c)$, $b=ab+bc+ca$, $c=-abc$

Case 1 : $c=0$ so $b=ab$ then $a=1$ or $b=0$

if $a=1$, we get $(a,b,c)=(1,-2,0)$

if $b=0$, we get $(a,b,c)=(0,0,0)$

Case 2 : $c\not=0$ so $ab=-1$ then $-1+c(a+b)=b$ ---[1]

Since $a=-(a+b+c)$, so $c=-2a+\frac{1}{a}$

substitute $c=-2a+\frac{1}{a}$ and $b=-\frac{1}{a}$ in [1]

we have $1+(a-\frac{1}{a})(2a-\frac{1}{a})=\frac{1}{a}$, then $(a-1)(2a^2(a+1)-1)=0$

If $2a^2(a+1)-1=0$, then $2a^3+2a^2-1=0$, by Rational root theorem, a $\not \in \mathbb{Q}$

so $a=1$, we get $(a,b,c)=(1,-1,-1)$

Answer : $P(x)=x^3$, $x^3+x^2-2x$, $x^3+x^2-x-1$