Question

If x = `(root3 (a + 1) + root3 (a - 1))/(root3 (a + 1) - root(3)(a - 1)`, prove that x³ – 3ax² + 3x – a = 0.

Answer 1

x = `(root(3)(a + 1) + root(3)(a - 1))/(root(3)(a + 1) - root(3)(a - 1)`
Applying componendo and dividendo,

`(x + 1)/(x - 1) = ((root(3)(a + 1) + root(3)(a - 1)) + (root(3)(a + 1) - root(3)(a - 1)))/((root(3)(a + 1) + root(3)(a - 1)) - (root(3)(a + 1) - root(3)(a - 1))`

`(x + 1)/(x - 1) = (2root(3)(a + 1))/(2root(3)(a - 1)`

⇒ `(x + 1)/(x - 1) = (root(3)(a + 1))/(root(3)(a - 1)`
Cubing both sides

`((x + 1)^3)/(x - 1)^3 = (a + 1)/(a - 1)`
Expand the cubes:

`(x^3 + 3x^2 + 3x + 1)/(x^3 - 3x^2 + 3x - 1) = (a + 1)/(a - 1)`

Apply componendo and dividendo again:

`((x^3 + 3x^2 + 3x + 1) + (x^3 - 3x^2 + 3x - 1))/((x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1)) = ((a + 1) + (a - 1))/((a + 1) - (a + 1))`

⇒ `(2x^3 + 6x)/(6x^2 + 2) = (2a)/2`

⇒ `(2(x^3 + 3x))/(2(3x^2 + 1))` = a

⇒ `(x^3 + 3x)/(3x^2 + 1) = a/(1)`
⇒ x³ + 3x = 3ax² + a
⇒ x³ – 3ax² + 3x – a = 0
Hence proved.