Factorise the expression: $x^{2} + \frac{y^{2}}{4} + \frac{z^{2}}{16} + xy + \frac{yz}{4} + \frac{zx}{2}$

(N/A) The given expression is of the form $a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca = (a + b + c)^{2}$.
Comparing the given expression $x^{2} + (\frac{y}{2})^{2} + (\frac{z}{4})^{2} + 2(x)(\frac{y}{2}) + 2(\frac{y}{2})(\frac{z}{4}) + 2(\frac{z}{4})(x)$ with the identity:
Here,$a = x$,$b = \frac{y}{2}$,and $c = \frac{z}{4}$.
Thus,the expression can be written as $(x + \frac{y}{2} + \frac{z}{4})^{2}$.
Therefore,the factors are $(x + \frac{y}{2} + \frac{z}{4})(x + \frac{y}{2} + \frac{z}{4})$.