Let $$P(x)=x^{3}+ax^{2}+bx+c$$ be a polynomial where $$a, b, c$$ are integers and $$c$$ is odd. Let $$p_{i}$$ be the value of $$P(x)$$ at $$x=i$$. Given that $$p_{1}^{3}+p_{2}^{3}+p_{3}^{3}=3p_{1}p_{2}p_{3}$$, find the value of $$p_{2}+2p_{1}-3p_{0}$$.

Let the cubic be $$P(x)=x^{3}+ax^{2}+bx+c$$ with integers $$a,b,c$$ and odd $$c$$.

Define $$p_i=P(i)$$ for $$i=0,1,2,3$$.
Explicitly

$$\begin{aligned} p_0 &= c,\\ p_1 &= 1+a+b+c,\\ p_2 &= 8+4a+2b+c,\\ p_3 &= 27+9a+3b+c. \end{aligned}$$

The given condition is

$$p_{1}^{3}+p_{2}^{3}+p_{3}^{3}=3p_{1}p_{2}p_{3}\,.$$

Recall the factorisation

$$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)\bigl(a^{2}+b^{2}+c^{2}-ab-bc-ca\bigr)\,.\;-(1)$$

Applying $$(1)$$ with $$a=p_1,\;b=p_2,\;c=p_3$$ gives

$$\bigl(p_1+p_2+p_3\bigr)\Bigl(p_1^{2}+p_2^{2}+p_3^{2}-p_1p_2-p_2p_3-p_3p_1\Bigr)=0.$$

Hence at least one of the following must hold:

Case 1: $$p_1+p_2+p_3=0$$

Sum the expressions for $$p_1,p_2,p_3$$:

$$p_1+p_2+p_3 = 36+14a+6b+3c.$$

Setting this to zero gives

$$14a+6b+3c=-36.\;-(2)$$

The left side of $$(2)$$ is the sum of two even numbers $$14a,6b$$ and an odd number $$3c$$ (because $$c$$ is odd). Therefore the left side is odd, whereas the right side $$-36$$ is even - a contradiction.
Thus Case 1 is impossible when $$c$$ is odd.

Case 2: $$p_1=p_2=p_3$$

Equating consecutive pairs:

$$\begin{aligned} p_1=p_2 &\Longrightarrow 1+a+b+c = 8+4a+2b+c \Longrightarrow b = -7-3a,\;-(3)\\[4pt] p_1=p_3 &\Longrightarrow 1+a+b+c = 27+9a+3b+c \Longrightarrow b = -13-4a.\;-(4) \end{aligned}$$

Solving $$(3)$$ and $$(4)$$ simultaneously:

$$-7-3a = -13-4a \;\;\Longrightarrow\;\; a=-6,$$

and then $$b = -7-3(-6)=11.$$

Therefore

$$a=-6,\qquad b=11,\qquad c \text{ any odd integer.}$$

Now evaluate the required expression

$$\begin{aligned} p_2 + 2p_1 - 3p_0 &= (8+4a+2b+c) + 2(1+a+b+c) - 3c \\[4pt] &= 10 + 6a + 4b \\[4pt] &= 10 + 6(-6) + 4(11) \\[4pt] &= 10 - 36 + 44 \\[4pt] &= 18. \end{aligned}$$

Notice that $$c$$ has cancelled out, so the value is the same for every odd $$c$$.

Hence the required value is 18.