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Question
Using the properties of proportion, solve the following equation for x; given `(x^3 + 3x)/(3x^2 + 1) = (341)/(91)`
Solution
`(x^3 + 3x)/(3x^2 + 1) = (341)/(91)`
Applying componendo and dividendo
`(x^3 + 3x + 3x^2 + 1)/(x^3 + 3x- 3x^2 - 1) = (341 + 91)/(341 - 91)`
⇒ `(x + 1)^3/(x - 1)^3 = (432)/(250)`
⇒ `(x + 1)^3/(x - 1)^3 = (216)/(125)`
⇒ `(x + 1)^3/(x - 1)^3 = (6/5)^3`
⇒ `(x + 1)/(x - 1) = (6)/(5)`
⇒ 6x – 6 = 5x + 5
⇒ 6x – 5x = 5 + 6
⇒ x = 11
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