Crazynomials??

Easy.

$a+b+c=3x$

$\text{Expanding this, we shall get}: a+b+c=x+x+x.$

$\text{The }a,\text{ }b, \text{ and } c \text{ all correspond to } x.$

$\text{Therefore, }x=a, x=b, \text{ and } x=c.$

$\text{Then solve the equation by making the right side equal to zero}.$

$x=a\rightarrow \color{#0000ff}{x-a=0}$

$x=b\rightarrow \color{#0000ff}{x-b=0}$

$x=c\rightarrow \color{#0000ff}{x-c=0}$

$\color{#ffffff}{\text{I am super awesome! lol}}$

$=(x-a)^3+(x-b)^3+(x-c)^3-3(x-a)(x-b)(x-c)$

$\color{#ffffff}{\downarrow\downarrow\downarrow\downarrow\downarrow}\downarrow\color{#ffffff}{\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow}\downarrow\color{#ffffff}{\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow}\downarrow\color{#ffffff}{\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow}\downarrow\color{#ffffff}{\downarrow\downarrow\downarrow\downarrow\downarrow}\downarrow\color{#ffffff}{\downarrow\downarrow\downarrow\downarrow\downarrow}\downarrow$

$=(\color{#ffffff}{\text{xx}}0\color{#ffffff}{\text{xx}})^3+(\color{#ffffff}{\text{xx}}0\color{#ffffff}{\text{x}})^3+\color{#ffffff}{x}(\color{#ffffff}{\text{xx}}0\color{#ffffff}{\text{xx}})^3-3(\color{#ffffff}{\text{xx}}0\color{#ffffff}{\text{xx}})(\color{#ffffff}{\text{xx}}0\color{#ffffff}{\text{xx}})(\color{#ffffff}{\text{xx}}0\color{#ffffff}{\text{xx}})$

$=0$

Since the question implies that it holds for every $a, b$ and $c$ that add up to $3x$ , choose $a=b=c=x$ .

(p+q+r)^{3} = p^{3} + q^{3} + r^{3} + 3(p+q+r)(pq+pr+qr) - 3pqr

then p = x-a ; q=x-b ; r=x-c

0^{3} = (x-a)^{3} + (x-b)^{3} + (x-c)^{3} + 0 - 3(x-a)(x-b)(x-c)

(x-a)^{3} + (x-b)^{3} + (x-c)^{3} - 3(x-a)(x-b)(x-c) = 0 yeah :)