3 – 27(x – y)3] ÷ (5Y – x) = Ax2 + Bxy + Cy2 ∴ A + B + C = 16Detailed Solution∶⇒ [8(x + y)3 – 27(x - y)3] ÷ (5y - x) = Ax2 + Bxy + Cy2⇒ [{2(x + y)}3 – {3(x – y)}3] ÷ (5y – x) = Ax2 + Bxy + Cy2⇒ {2(x + y)} – {3(x – y)} [{2(x + y)}2 + 2(x + y)3(x – y) + {3(x – y)}2] ÷ (5y – x) = Ax2 + Bxy + Cy2⇒ (5y – x) [4x2 + 4Y2 + 8xy + 6x2 – 6Y2 + 9x2 + 9y2 – 18xy] ÷ (5y – x) = Ax2 + Bxy + Cy2⇒ 19x2 – 10xy + 7y2 = Ax2 + Bxy + Cy2On comparing⇒ A = 19, B = -10 and C = 7⇒ (A + B + C) = 19 + 7 – 10 = 16

"> 3 – 27(x – y)3] ÷ (5Y – x) = Ax2 + Bxy + Cy2 ∴ A + B + C = 16Detailed Solution∶⇒ [8(x + y)3 – 27(x - y)3] ÷ (5y - x) = Ax2 + Bxy + Cy2⇒ [{2(x + y)}3 – {3(x – y)}3] ÷ (5y – x) = Ax2 + Bxy + Cy2⇒ {2(x + y)} – {3(x – y)} [{2(x + y)}2 + 2(x + y)3(x – y) + {3(x – y)}2] ÷ (5y – x) = Ax2 + Bxy + Cy2⇒ (5y – x) [4x2 + 4Y2 + 8xy + 6x2 – 6Y2 + 9x2 + 9y2 – 18xy] ÷ (5y – x) = Ax2 + Bxy + Cy2⇒ 19x2 – 10xy + 7y2 = Ax2 + Bxy + Cy2On comparing⇒ A = 19, B = -10 and C = 7⇒ (A + B + C) = 19 + 7 – 10 = 16

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If [8(x + y)3 – 27(x – y)3] ÷ (5y – x) = Ax2 + Bxy + Cy2, then the value of (A + B + C) is:

General Aptitude Algebra in General Aptitude 8 months ago

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Short trick∶Put x = y = 1 in [8(x + y)3 – 27(x – y)3] ÷ (5Y – x) = Ax2 + Bxy + Cy2 ∴ A + B + C = 16Detailed Solution∶⇒ [8(x + y)3 – 27(x - y)3] ÷ (5y - x) = Ax2 + Bxy + Cy2⇒ [{2(x + y)}3 – {3(x – y)}3] ÷ (5y – x) = Ax2 + Bxy + Cy2⇒ {2(x + y)} – {3(x – y)} [{2(x + y)}2 + 2(x + y)3(x – y) + {3(x – y)}2] ÷ (5y – x) = Ax2 + Bxy + Cy2⇒ (5y – x) [4x2 + 4Y2 + 8xy + 6x2 – 6Y2 + 9x2 + 9y2 – 18xy] ÷ (5y – x) = Ax2 + Bxy + Cy2⇒ 19x2 – 10xy + 7y2 = Ax2 + Bxy + Cy2On comparing⇒ A = 19, B = -10 and C = 7⇒ (A + B + C) = 19 + 7 – 10 = 16

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Posted on 22 Oct 2024, this text provides information on General Aptitude related to Algebra in General Aptitude. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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