d:[["$","title","0",{"children":"IIM Kozhikode - If x = [3]28 and y = [3]27, find the value of x + y - 1/x^2+xy+y^2. | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined Looking at this prob<=m, I need to find the value of x + y - 1x^2+xy+y^2 where x = [3]28 and y = [3]27.

Since y = [3]27 and 27 = 3^3, we >=t: y = [3]3^3 = 3

The expression x^2 + xy + y^2 in the denominator connects to the factorization: x^3 - y^3 = (x-y)(x^2+xy+y^2) This means: x^2 + xy + y^2 = x^3 - y^3x-y This is useful because we can calculate x^3 and y^3 easily!

Calculating the cube values: - x^3 = ([3]28)^3 = 28 - y^3 = ([3]27)^3 = 27 Substituting into our identity: x^2 + xy + y^2 = x^3 - y^3x-y = 28 - 27x-y = 1x-y

Substituting back into the original expression: x + y - 1x^2+xy+y^2 = x + y - 11x-y When we divide by a fraction, we multiply by its reciprocal: = x + y - (x-y) = x + y - x + y = 2y

Since y = 3: 2y = 2 3 = 6 Notice how we avoided calculating messy cube roots like [3]28! The al>=braic identity x^3 - y^3 = (x-y)(x^2+xy+y^2) made this prob<=m simp<=. Answer: 6"}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"IIM-K (BMSAT),QA,Algebra,Identities"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"property":"og:title","content":"IIM Kozhikode - If x = [3]28 and y = [3]27, find the value of x + y - 1/x^2+xy+y^2. | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined Looking at this prob<=m, I need to find the value of x + y - 1x^2+xy+y^2 where x = [3]28 and y = [3]27.

Since y = [3]27 and 27 = 3^3, we >=t: y = [3]3^3 = 3

Calculating the cube values: - x^3 = ([3]28)^3 = 28 - y^3 = ([3]27)^3 = 27 Substituting into our identity: x^2 + xy + y^2 = x^3 - y^3x-y = 28 - 27x-y = 1x-y

Substituting back into the original expression: x + y - 1x^2+xy+y^2 = x + y - 11x-y When we divide by a fraction, we multiply by its reciprocal: = x + y - (x-y) = x + y - x + y = 2y

Since y = 3: 2y = 2 3 = 6 Notice how we avoided calculating messy cube roots like [3]28! The al>=braic identity x^3 - y^3 = (x-y)(x^2+xy+y^2) made this prob<=m simp<=. Answer: 6"}],["$","meta","9",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/iimk-bmsat/2025/QA/10"}],["$","meta","10",{"property":"og:site_name","content":"AfterBoards | IIM Kozhikode Preparation"}],["$","meta","11",{"property":"og:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","meta","12",{"property":"og:type","content":"website"}],["$","meta","13",{"name":"twitter:card","content":"summary_large_image"}],["$","meta","14",{"name":"twitter:site","content":"afterboards.in"}],["$","meta","15",{"name":"twitter:title","content":"IIM Kozhikode - If x = [3]28 and y = [3]27, find the value of x + y - 1/x^2+xy+y^2. | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined Looking at this prob<=m, I need to find the value of x + y - 1x^2+xy+y^2 where x = [3]28 and y = [3]27.

Since y = [3]27 and 27 = 3^3, we >=t: y = [3]3^3 = 3

Calculating the cube values: - x^3 = ([3]28)^3 = 28 - y^3 = ([3]27)^3 = 27 Substituting into our identity: x^2 + xy + y^2 = x^3 - y^3x-y = 28 - 27x-y = 1x-y

Substituting back into the original expression: x + y - 1x^2+xy+y^2 = x + y - 11x-y When we divide by a fraction, we multiply by its reciprocal: = x + y - (x-y) = x + y - x + y = 2y

Since y = 3: 2y = 2 3 = 6 Notice how we avoided calculating messy cube roots like [3]28! The al>=braic identity x^3 - y^3 = (x-y)(x^2+xy+y^2) made this prob<=m simp<=. Answer: 6"}],"$L17","$L18","$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f","$L20"]

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