UPSC CDS 2 2021 Math Paper

Concept:The given expression is symmetric and homogeneous of degree zero.We can evaluate it by substituting a convenient value (e.g., $z = 0$) without changing its constant value.Explanation:Let $P = \frac{x^{4}}{(x^{2}-y^{2})(x^{2}-z^{2})} + \frac{y^{4}}{(y^{2}-x^{2})(y^{2}-z^{2})} + \frac{z^{4}}{(z^{2}-x^{2})(z^{2}-y^{2})}$.Because the expression is symmetric in $x, y, z$, its value does not depend on the specific variables (as long as no denominator becomes zero).Choose $z = 0$ to simplify. This is allowed because the expression is defined for general nonzero values.Substitute $z = 0$:First term: $\frac{x^{4}}{(x^{2}-y^{2})(x^{2}-0)} = \frac{x^{4}}{(x^{2}-y^{2})x^{2}} = \frac{x^{2}}{x^{2}-y^{2}}$.Second term: $\frac{y^{4}}{(y^{2}-x^{2})(y^{2}-0)} = \frac{y^{4}}{(y^{2}-x^{2})y^{2}} = \frac{y^{2}}{y^{2}-x^{2}} = -\frac{y^{2}}{x^{2}-y^{2}}$.Third term: $\frac{0}{(0-x^{2})(0-y^{2})} = 0$.Now sum the first two terms:$\frac{x^{2}}{x^{2}-y^{2}} - \frac{y^{2}}{x^{2}-y^{2}} = \frac{x^{2} - y^{2}}{x^{2} - y^{2}} = 1$.Thus, for $z=0$, $P = 1$.Because the expression is symmetric and of degree zero, its value is constant for all $x,y,z$ (excluding points where denominators vanish).Therefore, the original sum equals $1$.Answer:$1$ (Option C).