What happened to the squares?

Note that ${ a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }-3abc=(a+b+c)({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca)$ .

However, $a+b+c=0$ , so ${ a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }=3abc$

Thus, $abc=\frac { 216 }{ 3 } =\boxed { 72 }$

Let $a,b,c$ be the roots of cubic polynomial $F(x)$ :

$p=a+b+c=0 \\ q=ab+ac+bc\\ r=abc$

$F(x)=(x-a)(x-b)(x-c)=x^3-px^2+qx-r=x^3+qx-r$

$F(a)=0 \implies a^3+qa-r=0 \implies a^3=r-q a$

We have symmetric results for $b$ and $c$ :

$a^3+b^3+c^3=216 \implies (r-qa)+(r-qb)+(r-qc)=216 \implies 3r-(a+b+c)q=216 \implies 3r=216$

$r=\dfrac{216}{3}=72$

$\color{#20A900}{\boxed{\boxed{abc=72}}}$