$abc$ is the one

Using AM-GM inequality , we have:

$\begin{aligned} a^3(b^2+c)+b^3(c^2+a)+c^3(a^2+b) & = a^3b^2+a^3c + b^3c^2+b^3a+c^3a^2+c^3b \ge 6\sqrt [6] {(\color{#3D99F6}{abc})^9} = \boxed{6} \quad \quad \small \color{#3D99F6}{\text{Since }abc = 1} \end{aligned}$

Relevant wiki: Arithmetic Mean - Geometric Mean

From AM-GM we have: $\\ \quad { a }^{ 2 }+b\ge 2a\sqrt { b } ;\quad b^{ 2 }+c\ge 2b\sqrt { c } ;\quad c^{ 2 }+a\ge 2c\sqrt { a } \quad =>\\ =>\quad ({ a }^{ 2 }+b)({ b }^{ 2 }+c)({ c }^{ 2 }+a)\ge 8abc\sqrt { abc } =8\quad <=>\\ <=>\quad { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }+{ a }^{ 3 }{ b }^{ 2 }+{ a }^{ 2 }{ c }^{ 3 }+{ a }^{ 3 }{ c }+{ b }^{ 3 }{ c }^{ 2 }+{ b }^{ 3 }{ a }+{ b }{ c }^{ 3 }+abc\ge 8\quad <=>\\ <=>\quad 1+{ a }^{ 3 }{ b }^{ 2 }+{ a }^{ 2 }{ c }^{ 3 }+{ a }^{ 3 }{ c }+{ b }^{ 3 }{ c }^{ 2 }+{ b }^{ 3 }{ a }+{ b }{ c }^{ 3 }+1\ge 8\quad <=>\\ <=>\quad { a }^{ 3 }{ b }^{ 2 }+{ a }^{ 2 }{ c }^{ 3 }+{ a }^{ 3 }{ c }+{ b }^{ 3 }{ c }^{ 2 }+{ b }^{ 3 }{ a }+{ b }{ c }^{ 3 }\ge 6\quad <=>\\ <=>\quad { a }^{ 3 }({ b }^{ 2 }+c)+b^{ 3 }({ c }^{ 2 }+a)+{ c }^{ 3 }({ a }^{ 2 }+b)\ge 6.\\$

Equality holds when $a=b=c$ , hence the answer is 6 .