Question

Question from Componendo and Dividendo (Class 11th Maths):- $If\:y=\frac{(p+1)^\frac{1}{3}+(p-1)^\frac{1}{3} }{(p+1)^\frac{1}{3}-(p-1)^\frac{1}{3} }\\then\:find\:the\:value\:of\:y^3-3py^2+3y-p\:=?$

Answer 1

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• $\boxed{y^3 - 3y^2 +3y - p = 0}$

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• If $a:b = c:d,$ then

$\:\:\:\:\:\:\:\:\:\:\: \frac{a+b}{a-b} = \frac{c+d}{c-d}$

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Here,

$\:\:\:\:\:\:y = \frac{{(p+1)} ^{\frac{1}{3}} + {(p-1)} ^{\frac{1}{3}}}{{(p+1)} ^{\frac{1}{3}} - {(p-1)} ^{\frac{1}{3}}}$

Applying the above formula,

$\implies \frac{y+1} {y-1}$

$\:\:\:\:\:\:= \frac{[(p+1)^{\frac{1}{3}} + (p-1)^{\frac{1}{3}}]+[(p+1)^{\frac{1}{3}} - (p-1)^{\frac{1}{3}}]}{[(p+1)^{\frac{1}{3}} + (p-1)^{\frac{1}{3}}]-[(p+1)^{\frac{1}{3}} - (p-1)^{\frac{1}{3}}]}$

$\implies \frac{y+1} {y-1} = \frac{2(p+1)^{\frac{1}{3}} }{2(p-1)^{\frac{1}{3}}}$

$\implies \frac{(y+1)^3}{(y-1)^3} = \frac{p+1}{p-1}$

Applying the above formula,

$\implies \frac{(y+1)^3+(y-1)^3}{(y+1)^3-(y-1)^3} = \frac{(p+1)+(p-1)}{(p+1)-(p-1)}$

$\implies \frac{2(y^3 + 3y)}{2(3y^2 + 1)} = \frac{2p}{2}$

$\implies \frac{y^3 + 3y}{3y^2 + 1} = \frac{p}{1}$

$\implies (y^3 + 3y) = p(3y^2 + 1)$

$\implies y^3 + 3y = 3py^2 + p$

$\implies y^3 - 3py^2 + 3y - p = 0$

Answer 2

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

⇒  y3 + 1 + 3y2 + 3y/y3 – 1 – 3y2 + 3y = p + 1/p – 1

Applying componendo and dividendo

⇒ y3 + 1 + 3y2 + 3y + y3 – 1 – 3y2 + 3y/y3 + 1 + 3y2 + 3y – y3 + 1 + 3y2 – 3y

=  p + 1 + p – 1/p + 1 – p + 1

⇒ 2y3 + 6y /6y2 + 2 = 2p/2

⇒ 2(y3 + 3y)/2(3y2 + 1) = p

⇒ y3 + 3y = 3py2 + p

⇒ y3 – 3py2 + 3y – p = 0.

Hence  proved.