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प्रश्न
If x = `(root(3)(a + 1) + root(3)(a - 1))/(root(3)(a + 1) - root(3)(a - 1)`,prove that :
x³ – 3ax² + 3x – a = 0
उत्तर
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)`
Again apply onendo and dividendo,
`((x + 1)^2 + (x - 1)^3)/((x + 1)^3 - (x - 1)^3) = (a + 1 + a - 1)/(a + 1 - a + 1)`
⇒ `(2(x^3 + 3x))/(2(3x^2 + 1)) = (2a)/(2)`
⇒ `(x^3 + 3x)/(3x^2 + 1) = a/(1)`
⇒ x3 + 3x = 3ax2 + a
⇒ x3 – 3ax2 + 3x – a = 0
Hence proved.
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