## Task 18/75

Determine all triples of real solutions $(x ; y ; z)$ of the equation

$$
x^{2}+y^{4}+z^{6}+14=2 x+4 y^{2}+6 z^{3}
$$
By rearrangement, the given equation becomes

$$
x^{2}-2 x+1-y^{4}-4 y+4+z^{6}-6 z+9=0 \rightarrow(x-1)^{2}+\left(y^{2}-2\right)^{2}+\left(z^{3}-3\right)^{2}=0
$$

A sum of squares of real numbers is zero if and only if each square is zero. Therefore, it follows that

$$
x=1 \quad ; \quad y= \pm \sqrt{2} \quad ; \quad z = \sqrt[3]{3}
$$

Thus, there exist the two solution triples $(1 ; \sqrt{2} ; \sqrt[3]{3})$ and $(1 ;-\sqrt{2} ; \sqrt[3]{3})$.