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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
Solution
`"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.
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