20:[["$","title","0",{"children":"BMSAT Kozhikode 2025 - If x = [3]28 and y = [3]27, find the value of x + y - 1/x^2+xy+y^2. | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined Since y = [3]27 and 27 = 3^3: y = [3]3^3 y = 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) Therefore: x^2 + xy + y^2 = x^3 - y^3x-y

Since x = [3]28 and y = [3]27: x^3 = ([3]28)^3 = 28 y^3 = ([3]27)^3 = 27

Substituting: x^2 + xy + y^2 = x^3 - y^3x-y x^2 + xy + y^2 = 28 - 27x-y x^2 + xy + y^2 = 1x-y

The original expression becomes: x + y - 1x^2+xy+y^2 = x + y - 11x-y = x + y - (x-y) = x + y - x + y = 2y

Since y = 3: 2y = 2 3 = 6"}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"BMSAT Kozhikode,QA,Algebra,Identities"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"name":"mobile-web-app-capable","content":"yes"}],["$","meta","8",{"name":"apple-mobile-web-app-title","content":"AfterBoards"}],["$","meta","9",{"name":"apple-mobile-web-app-status-bar-style","content":"default"}],["$","meta","10",{"property":"og:title","content":"BMSAT Kozhikode 2025 - If x = [3]28 and y = [3]27, find the value of x + y - 1/x^2+xy+y^2. | PYQs + Solutions"}],["$","meta","11",{"property":"og:description","content":"undefined Since y = [3]27 and 27 = 3^3: y = [3]3^3 y = 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) Therefore: x^2 + xy + y^2 = x^3 - y^3x-y

Since x = [3]28 and y = [3]27: x^3 = ([3]28)^3 = 28 y^3 = ([3]27)^3 = 27

Substituting: x^2 + xy + y^2 = x^3 - y^3x-y x^2 + xy + y^2 = 28 - 27x-y x^2 + xy + y^2 = 1x-y

The original expression becomes: x + y - 1x^2+xy+y^2 = x + y - 11x-y = x + y - (x-y) = x + y - x + y = 2y

Since y = 3: 2y = 2 3 = 6"}],["$","meta","12",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/iimk-bmsat/2025/QA/10"}],["$","meta","13",{"property":"og:site_name","content":"AfterBoards | BMSAT Kozhikode 2025 Preparation"}],["$","meta","14",{"property":"og:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","meta","15",{"property":"og:type","content":"website"}],["$","meta","16",{"name":"twitter:card","content":"summary_large_image"}],["$","meta","17",{"name":"twitter:site","content":"afterboards.in"}],["$","meta","18",{"name":"twitter:title","content":"BMSAT Kozhikode 2025 - If x = [3]28 and y = [3]27, find the value of x + y - 1/x^2+xy+y^2. | PYQs + Solutions"}],["$","meta","19",{"name":"twitter:description","content":"undefined Since y = [3]27 and 27 = 3^3: y = [3]3^3 y = 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) Therefore: x^2 + xy + y^2 = x^3 - y^3x-y

Since x = [3]28 and y = [3]27: x^3 = ([3]28)^3 = 28 y^3 = ([3]27)^3 = 27

Substituting: x^2 + xy + y^2 = x^3 - y^3x-y x^2 + xy + y^2 = 28 - 27x-y x^2 + xy + y^2 = 1x-y

The original expression becomes: x + y - 1x^2+xy+y^2 = x + y - 11x-y = x + y - (x-y) = x + y - x + y = 2y

Since y = 3: 2y = 2 3 = 6"}],["$","meta","20",{"name":"twitter:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","link","21",{"rel":"shortcut icon","href":"https://www.afterboards.in/favicon.ico","type":"image/x-icon","sizes":"48x48"}],"$L2e","$L2f","$L30","$L31","$L32","$L33","$L34","$L35"]

If $x = \sqrt[3]{28}$ and $y = \sqrt[3]{27}$, find the value of $x + y - \frac{1}{x^2+xy+y^2}$.

Keyboard Shortcuts