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प्रश्न
If y = `((p + 1)^(1/3) + (p - 1)^(1/3))/((p + 1)^(1/3) - (p - 1)^(1/3)` find that y3 - 3py2 + 3y - p = 0.
उत्तर
We have
`y/(1) = ((p + 1)^(1/3) + (p - 1)^(1/3))/((p + 1)^(1/3) - (p - 1)^(1/3)`
Applying componendo and dividendo
`(y + 1)/(y - 1) = ((p + 1)^(1/3) + (p - 1)^(1/3) + (p + 1)^(1/3) - (p - 1)^(1/3))/((p + 1)^(1/3) + (p - 1)^(1/3) - (p + 1)^(1/3) + (p - 1)^(1 /3)`
⇒ `(y + 1)/(y - 1) = (2(p + 1)^(1/3))/(2(p - 1)^(1/3))`
Cubing both side
`((y + 1)^3)/((y - 1)^3) = (p + 1)/(p - 1)`
⇒ `(y^3 + 1 + 3y^2 + 3y)/(y^3 - 1 - 3y^2 + 3y) = (p + 1)/(p - 1)`
Again applying componendo and dividendo
⇒ `(y^3 + 1 + 3y^2 + 3y + y^3 - 1 - 3y^2 + 3y)/(y^3 + 1 + 3y^2 + 3y - y^3 + 1 3y^2 - 3y)`
= `(p + 1 + p - 1)/(p + 1 - p + 1)`
⇒ `(2y^3 + 6y)/(6y^2 + 2) = (2p)/(2)`
⇒ `(2(y^3 + 3y))/(2(3y^2 + 1)) = p`
⇒ y3 + 3y = 3py2 + p
⇒ y3 - 3py2 + 3y - p = 0.
Hence proved.
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