Find the value of function at $\,x=5$ If $\,f(x)\,$ is a non-constant polynomial of $\,x\,$ such that $\,f\left(x^3\right)-f\left(x^3-2\right)=f^2\left(x\right)+12\,$ is true for all $\,x\,$ then find the value of $\,f\left(5\right).\,$

HINT: Substitute $\,f(x) = ax^3 + bx^2 + cx + d$ into your equation and collect coefficients of the different powers of $x$. 
Solve obtained system of linear equations, thus getting an explicit expression for  $f$.
Substitute $\,x = 5\,$ into obtained equation and get the answer.

How do we know that $f$ is the third-order polynomial?    
On the left we have expression involving $\;f\left(x^3\right),\,$ and on the right $\;f^2\left(x\right).\,$
Assuming $\,f\,$ is a polynomial of degree $\,n\in\mathbb R,\,$ we conclude that
$\,3\left(n-1\right) =  2n.\,$
Therefore we conclude that polynomial $\,f\,$ is of order $3$, i.e. 
$\;f(x) = ax^3 + bx^2 + cx + d.\,$

EDIT: Upon request of @AjaySharma I provide explicit solution:
$$
f(x) = a x^3 + b x^2 + c x + d 
\implies 
\begin{cases}
f\left(x^3\right) =  a x^9 + b x^6 + c x^3 + d 
\\
f\left(x^3 - 2\right) =  a \left(x^3-2\right)^3 + b \left(x^3-2\right)^2 + c \left(x^3-2\right) + d 
\end{cases}
$$
Thus the difference 
$$
\begin{aligned}
f\left(x^3\right)-f\left(x^3-2\right) 
& = a \left(\left(x^3\right)^3 - \left(x^3-2\right)^3 \right)
+ b \left( \left(x^3\right)^2 - \left(x^3-2\right)^2 \right) 
+ c \left( \left(x^3\right) - \left(x^3-2\right) \right) 
\\ & =
2a \left(\left(x^3\right)^2 + \left(x^3\right) \left(x^3-2\right)+ \left(x^3-2\right)^2 \right) + 
2b \left( 2x^3-2 \right)  + 2c 
\\ & =
2a \left({x^6} + {x^6} - {2x^3} + {x^6} - {4x^3} + 4 \right) + 
4b \left( x^3-1 \right)  + 2c 
\\ & =
6 a x^6 + \left(4 b - 12 a\right) x^3 + 8 a - 4 b + 2 c
\end{aligned}
$$
On the other hand, 
$$
\begin{aligned}
f^2\left(x\right) & = \left(a x^3 + b x^2 + c x + d \right)^2 = 
\left(a x^3 + b x^2 + c x + d \right)\cdot \left(a x^3 + b x^2 + c x + d \right)
= \\
& = a^2x^6 + \left(2ab\right)x^5 + \left(b^2 + 2ac\right)x^4 + \left(2ad + 2bc\right)x^3 + \left(c^2 + 2bd\right)x^2 + \left(2cd\right)x + d^2
\end{aligned}
$$
Therefore  $\qquad f\left(x^3\right)-f\left(x^3-2\right)  = f^2\left(x\right)+12
\implies $
$$
\begin{aligned}
\implies& 
\ 6 a x^6 + \left(4 b - 12 a\right) x^3 + 8 a - 4 b + 2 c = 
\\ 
=&\ a^2x^6 + \left(2ab\right)x^5 + \left(b^2 + 2ac\right)x^4 + \left(2ad + 2bc\right)x^3 + \left(c^2 + 2bd\right)x^2 + \left(2cd\right)x + d^2 + 12 
\end{aligned}
$$
Let us gather coefficients by powers of $x$:
$$
\begin{aligned}
&x^6:& a^2 &= 6 a &\implies& & a &= 6 \\
&x^5:& 2ab &= 0 &\implies& & b &= 0  \\
&x^4:& b^2+2ac &= 0  &\implies& & c &= 0 \\
&x^3:& 2ad + 2bc &= 4 b - 12 a  &\implies& & d &= -6 \\
&x^2:& c^2 + 2bd &= 0 \\
&x^1:& 2cd &= 0 \\
&x^0:& d^2 + 12 &= 8a - 4b + 2c
\end{aligned}
$$
Thus we get 
$$
\bbox[5pt, border:2.5pt solid #FF0000]{f\left(x\right) = 6x^3-6}
$$
and so $\,f\left(5\right) = 744 .\,$
