Question

Suppose a, b are positive real numbers such that a√a+b√b=91 and a√b +b√a=84.Find a+b .
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Answer 1

Step-by-step explanation:

Given Suppose a, b are positive real numbers such that a√a+b√b=91 and a√b +b√a=84. Find a+b .  

• Given a√a + b√b = 91
•     So a√b + b√a = 84
• Now √ab (√a + √b) = 84
• Let m = √a and n = √b
• So (√a)^3 + (√b)^3 = 91
• So m^3 + n^3 = 91 -----------1
• So mn (m + n) = 84------------2
• Multiply both sides by 3 we get
•      3mn(m + n) = 84 x 3
•      3 mn (m + n) = 252 ---------3
• Adding equation 1 and 3 we get
• So m^3 + n^3 + 3mn (m + n) = 343
• Or m^3 + n^3 + 3mn (m + n) = 7^3
•      Or (m + n)^3 = 7^3
•      Or m + n = 7
• Consider m^3 + n^3 = 91
•       (m + n) (m^2 + n^2 – mn) = 91
•    (m + n) (m^2 + n^2) – mn (m + n) = 91
•       7 (m^2 + n^2) – 84 = 91 (since mn(m + n) = 84)
•      7(m^2 + n^2) = 175
• Or m^2 + n^2 = 25
•                 Now √a = m and √b = n
•               Or a = m^2 and b = n^2
•                  Or m^2 + n^2 = 25
•                           Or a + b = 25

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https://brainly.in/question/3025106