5. Solve the equation $\sqrt[3]{15 x+1-x^{2}}+\sqrt[3]{x^{2}-15 x+27}=4$.
Answer: $x=0, x=2, x=13, x=15$.

Let's make a substitution: $a=\sqrt[3]{15 x+1-x^{2}}, b=\sqrt[3]{x^{2}-15 x+27}$. The equation is equivalent to the system

$$
\left\{\begin{array} { l } 
{ a + b = 4 , } \\
{ a ^ { 3 } + b ^ { 3 } = 2 8 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ a + b = 4 , } \\
{ ( a + b ) ( a ^ { 2 } - a b + b ^ { 2 } ) = 2 8 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ a + b = 4 , } \\
{ ( a + b ) ^ { 2 } - 3 a b = 7 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a+b=4 \\
a b=3
\end{array}\right.\right.\right.\right.
$$

which in turn is equivalent to the combination of two systems:

$\left\{\begin{array}{l}\sqrt[3]{15 x+1-x^{2}}=1, \\ \sqrt[3]{x^{2}-15 x+27}=3\end{array}\right.$ or $\left\{\begin{array}{l}\sqrt[3]{15 x+1-x^{2}}=3, \\ \sqrt[3]{x^{2}-15 x+27}=1 .\end{array}\right.$

The solution of the first system reduces to solving the equation: $x^{2}-15 x=0 \Rightarrow x=0, x=15$. The solution of the second system reduces to solving the equation $x^{2}-15 x+26=0 \Rightarrow x=2, x=13$.