Question

Using the properties of proportion, solve the following equation for x; given `(x^3 + 3x)/(3x^2 + 1) = (341)/(91)`

Answer 1

`(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