Piece of cake

@Gurīdo Cuong , please tell me whether I did same as yours?

By AM-GM, we have

$\large{\frac { { a }^{ 3 } }{ 2 } +\frac { { a }^{ 3 } }{ 2 } +\frac { { b }^{ 2 } }{ 3 } +\frac { { b }^{ 2 } }{ 3 } +\frac { { b }^{ 2 } }{ 3 } +\frac { c }{ 6 } +\frac { c }{ 6 } +\frac { c }{ 6 } +\frac { c }{ 6 } +\frac { c }{ 6 } +\frac { c }{ 6 } \ge 11\sqrt [ 11 ]{ \frac { { \left( abc \right) }^{ 6 } }{ { 2 }^{ 2 }\times { 3 }^{ 3 }\times { 6 }^{ 6 } } } \\ 10\left( { a }^{ 3 }+{ b }^{ 2 }+c \right) \ge 110\sqrt [ 11 ]{ \frac { 1 }{ { 2 }^{ 8 }\times { 3 }^{ 9 } } } \approx 27.045}$

This means the minimum value of $10\left( { a }^{ 3 }+{ b }^{ 2 }+c \right)$ is approximately 27. Equality holds when $\frac { { a }^{ 3 } }{ 2 } =\frac { { b }^{ 2 } }{ 3 } =\frac { c }{ 6 }$

We assume that $a=l$ and $b=k$ when the equality holds

Using AM-GM we have $a^3+2l^3\geq 3al^2$ and $b^2+k^2\geq 2bk$

So
$a^3+b^2+c\geq 3al^2+2bk+c-2l^3-k^2\geq 3\sqrt[3]{6l^2k}-2l^3-k^2$ Now k and l must satisfy $\begin{cases} a=l\\b=k\\3al^2=2bk=c\\lkc=1\end{cases}$ Solving and we get $\begin{cases} l=\sqrt[11]{\frac{2}{27}}\\k=\sqrt{\frac{3}{2}\sqrt[11]{(\frac{2}{27})^3}}\\c=3\sqrt[11]{(\frac{2}{27})^3}\end{cases}$ So the minimum of the expression is $\approx 27.045...$ which is $27$ when rounded to the nearest integer