UPSC CDS 2 2022 Math Paper

Concept:The HCF of two algebraic expressions is the product of the common factors with the smallest exponent.Use the identities $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})$ and $a^2 - b^2 = (a-b)(a+b)$.Explanation:Step 1: Factorise the first expression $x^8 - y^8$.$x^8 - y^8 = (x^4)^2 - (y^4)^2 = (x^4 - y^4)(x^4 + y^4)$.$x^4 - y^4 = (x^2 - y^2)(x^2 + y^2) = (x-y)(x+y)(x^2 + y^2)$.So $x^8 - y^8 = (x-y)(x+y)(x^2 + y^2)(x^4 + y^4)$.Step 2: Factorise the second expression $x^7 - y^7 + x^5y^2 - x^2y^5$.Group terms: $(x^7 + x^5y^2) - (y^7 + x^2y^5) = x^5(x^2 + y^2) - y^5(x^2 + y^2)$.Factor out $(x^2 + y^2)$: $(x^2 + y^2)(x^5 - y^5)$.Now $x^5 - y^5 = (x-y)(x^4 + x^3y + x^2y^2 + xy^3 + y^4)$.Thus the second expression becomes $(x^2 + y^2)(x-y)(x^4 + x^3y + x^2y^2 + xy^3 + y^4)$.Step 3: Identify common factors from both factorised forms.First: $(x-y)(x+y)(x^2+y^2)(x^4+y^4)$.Second: $(x-y)(x^2+y^2)(x^4 + x^3y + x^2y^2 + xy^3 + y^4)$.Common factors are $(x-y)$ and $(x^2+y^2)$.Hence HCF $= (x-y)(x^2+y^2) = (x-y)(x^2+y^2)$.Step 4: Expand the product to match the given options.$(x-y)(x^2+y^2) = x \cdot x^2 + x \cdot y^2 - y \cdot x^2 - y \cdot y^2 = x^3 + xy^2 - x^2y - y^3$.Rearrange: $x^3 - y^3 - x^2y + xy^2$.This matches option C.Answer:Option C: $(x^3 - y^3 - x^2y + xy^2)$