UPSC CDS 2 2023 Math Paper

Concept:The expression can be rewritten using the identity $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.Explanation:First, express the given polynomial $3\sqrt{3}x^{3}+2\sqrt{2}y^{3}-18xy+6\sqrt{6}$ in the form of cubes and product.Note that $3\sqrt{3}x^{3}=(\sqrt{3}x)^{3}$, $2\sqrt{2}y^{3}=(\sqrt{2}y)^{3}$, and $6\sqrt{6}=(\sqrt{6})^{3}$.Also, $18xy=3 \cdot \sqrt{3}x \cdot \sqrt{2}y \cdot \sqrt{6}$.Therefore, the expression becomes $(\sqrt{3}x)^{3}+(\sqrt{2}y)^{3}+(\sqrt{6})^{3}-3(\sqrt{3}x)(\sqrt{2}y)(\sqrt{6})$.Let $a=\sqrt{3}x$, $b=\sqrt{2}y$, $c=\sqrt{6}$.Applying the identity $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$, we get the factorisation:$(\sqrt{3}x+\sqrt{2}y+\sqrt{6})\big[(\sqrt{3}x)^{2}+(\sqrt{2}y)^{2}+(\sqrt{6})^{2}-(\sqrt{3}x)(\sqrt{2}y)-(\sqrt{2}y)(\sqrt{6})-(\sqrt{6})(\sqrt{3}x)\big]$.Simplify the second bracket: $3x^{2}+2y^{2}+6-\sqrt{6}xy-\sqrt{12}y-\sqrt{18}x$.Thus one factor is $3x^{2}+2y^{2}+6-\sqrt{6}xy-\sqrt{12}y-\sqrt{18}x$.Answer:The required factor is $3x^{2}+2y^{2}+6-\sqrt{6}xy-\sqrt{12}y-\sqrt{18}x$, which corresponds to option C.