3 + (√b)3 = 32a√b + b√a = 31√a√b(√a + √b) = 31Suppose √a = x and √b = yx3 + Y3 = 32 ---- (1)xy(x + y) = 31 ---- (2)(x + y)3 = x3 + y3 + 3xy(x + y)(x + y)3 = 32 + 3 × 31 = 125 = 53(x + y) = 5x3 + y3 = (x + y)(x2 + y2 – xy) = 32(x + y)(x2 + y2 – xy(x + y) = 325(x2 + y2) – 31 = 32(x2 + y2) = 63/5(a + b) = 63/55(a + b)/7 = (5/7) × (63/5) = 9

"> 3 + (√b)3 = 32a√b + b√a = 31√a√b(√a + √b) = 31Suppose √a = x and √b = yx3 + Y3 = 32 ---- (1)xy(x + y) = 31 ---- (2)(x + y)3 = x3 + y3 + 3xy(x + y)(x + y)3 = 32 + 3 × 31 = 125 = 53(x + y) = 5x3 + y3 = (x + y)(x2 + y2 – xy) = 32(x + y)(x2 + y2 – xy(x + y) = 325(x2 + y2) – 31 = 32(x2 + y2) = 63/5(a + b) = 63/55(a + b)/7 = (5/7) × (63/5) = 9

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Let a and b be two positive real numbers such that a√a + b√b = 32 and a√b + b√a = 31. What is the value of 5(a + b)/7?

General Aptitude Algebra in General Aptitude 11 months ago

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a√a + b√b = 32(√a)3 + (√b)3 = 32a√b + b√a = 31√a√b(√a + √b) = 31Suppose √a = x and √b = yx3 + Y3 = 32 ---- (1)xy(x + y) = 31 ---- (2)(x + y)3 = x3 + y3 + 3xy(x + y)(x + y)3 = 32 + 3 × 31 = 125 = 53(x + y) = 5x3 + y3 = (x + y)(x2 + y2 – xy) = 32(x + y)(x2 + y2 – xy(x + y) = 325(x2 + y2) – 31 = 32(x2 + y2) = 63/5(a + b) = 63/55(a + b)/7 = (5/7) × (63/5) = 9

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Posted on 07 Nov 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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