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Question
Given that `(a^3 + 3ab^2)/(b^3 + 3a^2b) = (63)/(62)`. Using componendo and dividendo find a: b.
Solution
Given that `(a^3 + 3ab^2)/(b^3 + 3a^2b) = (63)/(62)`
By componendo and dividendo
`(a^3 + 3ab^2 + b^3 + 3a^2b)/(a^3 + 3ab^2 - b^3 - 3a^2b) = (63 + 62)/(63 - 62) = (125)/(1)`
⇒ `(a + b)^3/(a b)^3 - (5/1)^3`
⇒ `(a + b)/(a - b)` = 5
⇒ a + b = 5a – 5b
⇒ 5a – a – 5b – b = 0
⇒ 4a – 6b = 0
⇒ 4a = 6b
⇒ `a/b = (6)/(4)`
⇒ `a/b = (3)/(2)`
a : b = 3 : 2.
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