Question

If x = `(root (3)("m + 1") + root (3)("m - 1"))/(root (3)("m + 1") + root (3)("m - 1")` then prove that x³ - 3mx² + 3x = m

Answer 1

`"x"/1 = (root (3)("m + 1") + root (3)("m - 1"))/(root (3)("m + 1") + root (3)("m - 1")`

Applying componendo and dividendo 

`("x"+ 1) /("x" - 1) = (root (3)("m + 1") + root (3)("m - 1") + root (3)("m + 1") - root (3)("m - 1"))/ (root (3)("m + 1") + root (3)("m - 1") - root (3)("m + 1") + root (3)("m - 1")`

`=> ("x"+ 1) /("x" - 1) = (2 root (3)("m" + 1))/(2 root (3)("m" - 1))`

Cubing both sides 

`=> ("x" + 1)^3/("x - 1")^3 = (8 ("m + 1"))/(8("m - 1"))`

`=> ("x"^3 + 3"x"^2 +3"x" +1)/("x"^3 - 3"x"^2 + 3"x" -1) = ("m + 1")/("m - 1")`

⇒  (m - 1) (x³ + 3x² + 3x + 1) = (m + 1 )(x³ - 3x² + 3x - 1) 

⇒ mx³ + 3mx² + 3mx + m - x³ - 3x² - 3x - 1 - mx³ - 3mx² + 3mx - m + x³ - 3x² + 3x -1

⇒ 6mx² + 2m - 2x³ - 6x = 0 

⇒ 3mx²+ m - x³ - 3x  = O 

⇒  x³ - 3mx² + 3x = m 

Hence Proved.