An algebra problem by Wildan Bagus Wicaksono

$\begin{aligned} x &= \sqrt[3]5 \left(\sqrt[3]{\frac{16}{25}} - \sqrt[3]{\frac{4}{25}} + \sqrt[3]{\frac{1}{25}} \right)^{-1} \\ &= \frac {\sqrt[3]5}{\sqrt[3]{\frac{16}{25}} - \sqrt[3]{\frac{4}{25}} + \sqrt[3]{\frac{1}{25}}} \\ &= \frac {\sqrt[3]5 \sqrt[3]{25}}{\sqrt[3]{16} - \sqrt[3]4 + 1} \\ &= \frac {5(\sqrt[3]4 + 1)}{(\sqrt[3]4 + 1)((\sqrt[3]4)^2 - \sqrt[3]4 + 1)} \\ &= \frac {5(\sqrt[3]4 + 1)}{(\sqrt[3]4)^3 + 1} \\ &= \frac {5(\sqrt[3]4 + 1)}5 \\ &= \sqrt[3]4 + 1 \end{aligned}$

$\implies \lceil x \rceil = \lceil \sqrt[3]4 + 1 \rceil = \boxed{3}$

$=\sqrt [ 3 ]{ 5 } { \left( \sqrt [ 3 ]{ \frac { 16 }{ 25 } } -\sqrt [ 3 ]{ \frac { 4 }{ 25 } } +\sqrt [ 3 ]{ \frac { 1 }{ 25 } } \right) }^{ -1 }\\ =\sqrt [ 3 ]{ 5 } { \left( \frac { 1 }{ \sqrt [ 3 ]{ 25 } } \left\{ \sqrt [ 3 ]{ 16 } -\sqrt [ 3 ]{ 4 } +1 \right\} \right) }^{ -1 }$

Let $y=\sqrt [ 3 ]{ 16 } -\sqrt [ 3 ]{ 4 } +1$ .

$y=\sqrt [ 3 ]{ 16 } -\sqrt [ 3 ]{ 4 } +1\quad \\ \sqrt [ 3 ]{ 4 } y=4-\sqrt [ 3 ]{ 16 } +\sqrt [ 3 ]{ 4 } \\ -----------------+\\ \left( 1+\sqrt [ 3 ]{ 4 } \right) y=5\\ y=\frac { 5 }{ 1+\sqrt [ 3 ]{ 4 } }$

$x=\sqrt [ 3 ]{ 5 } { \left( \frac { 1 }{ \sqrt [ 3 ]{ 25 } } \cdot \frac { 5 }{ 1+\sqrt [ 3 ]{ 4 } } \right) }^{ -1 }\\ x=\frac { \sqrt [ 3 ]{ 5 } \cdot \sqrt [ 3 ]{ 25 } \left( 1+\sqrt [ 3 ]{ 4 } \right) }{ 5 } \\ x=1+\sqrt [ 3 ]{ 4 } \\ x=1+1,...\\ x=2,...\\ \left\lceil x \right\rceil =3$

So, $\left\lceil x \right\rceil =\boxed { 3 }$ .