Question

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Answer 1

$\underline{\underline{\Huge{\textbf{Question:}}}}$

$If\: \: p\: cos^{3}\, \theta + 3p\: cos\, \theta\: sin^{2}\: \theta= m\: \: and\: \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: p\: sin^{3}\, \theta + 3p\: cos^{2}\, \theta\: sin\: \theta= n$

$Prove\: that \left(m+n\right)^{\frac{2}{3}}-\left(m-n\right)^{\frac{2}{3}}= 2p^{\frac{2}{3}}$



$\underline{\underline{\Huge{\textbf{Solution:}}}}$

$\underline{\Large{\textsf{Step-1:}}}$
Find m+n

$m+n = p\: cos^{3}\, \theta + 3p\: cos\, \theta\: sin^{2}\: \theta + p\: sin^{3}\, \theta + 3p\: sin^{2}\, \theta\: cos\: \theta$

$\implies m+n = p\left(cos^{3}\, \theta + 3cos\, \theta\: sin^{2}\: \theta + sin^{3}\, \theta + 3p\: sin^{2}\, \theta\: cos\: \theta\right)$————[1]



$\underline{\Large{\textsf{Step-2:}}}$
Rewrite [1] Using the Identity
$\bigstar \: \: \mathbf{a^{3}+3a^{2}b+3ab^{2}+b^{3} = \left(a+b\right)^{3}}$

$\implies m+n = p\left(sin\, \theta + cos\, \theta\right)^{3}$



$\underline{\Large{\textsf{Step-3:}}}$
Find m-n

$m-n = p\: cos^{3}\, \theta + 3p\: cos\, \theta\: sin^{2}\: \theta - \left(p\: sin^{3}\, \theta + 3p\: sin^{2}\, \theta\: cos\: \theta\right)$

$\implies m-n = p\: cos^{3}\, \theta + 3p\: cos\, \theta\: sin^{2}\: \theta - p\: sin^{3}\, \theta - 3p\: sin^{2}\, \theta\: cos\: \theta$

$\implies m-n = p\left(cos^{3}\, \theta + 3 cos\, \theta\: sin^{2}\: \theta - sin^{3}\, \theta - 3sin^{2}\, \theta\: cos\: \theta\right)$————[2]



$\underline{\Large{\textsf{Step-3:}}}$
Rewrite [2] Using the Identity
$\bigstar \: \: \mathbf{a^{3}-3a^{2}b+3ab^{2}-b^{3} = \left(a-b\right)^{3}}$

$\implies m-n = p\left(cos\, \theta - sin\, \theta\right)^{3}$



$\underline{\Large{\textsf{Step-4:}}}$
Find $\left(m+n\right)^{\frac{2}{3}}$

$\implies \left(m+n\right)^{\frac{2}{3}} = p^{\frac{2}{3}}\left(\left(sin\, \theta + cos\, \theta\right)^{\cancel{3}}\right)^{\frac{2}{\cancel{3}}}$

$\implies \left(m+n\right)^{\frac{2}{3}} = p^{\frac{2}{3}}\left(sin\, \theta + cos\, \theta\right)^{2}$ ————[3]



$\underline{\Large{\textsf{Step-5:}}}$
Find $\left(m-n\right)^{\frac{2}{3}}$

$\implies \left(m-n\right)^{\frac{2}{3}} = p^{\frac{2}{3}}\left(\left(cos\, \theta - sin\, \theta\right)^{\cancel{3}}\right)^{\frac{2}{\cancel{3}}}$

$\implies \left(m-n\right)^{\frac{2}{3}} = p^{\frac{2}{3}}\left(cos\, \theta - sin\, \theta\right)^{2}$ ————[4]



$\underline{\Large{\textsf{Step-6:}}}$
Add [3] & [4]

$\implies \left(m+n\right)^{\frac{2}{3}}+ \left(m-n\right)^{\frac{2}{3}} =p^{\frac{2}{3}}\left(sin\, \theta + cos\, \theta\right)^{2} + p^{\frac{2}{3}}\left(cos\, \theta - sin\, \theta\right)^{2}$

$\implies \left(m+n\right)^{\frac{2}{3}}+ \left(m-n\right)^{\frac{2}{3}} = p^{\frac{2}{3}} \left[\left(sin\, \theta + cos\, \theta\right)^{2} + \left(cos\, \theta - sin\, \theta\right)^{2}\right]$



$\underline{\Large{\textsf{Step-7:}}}$
Using the Identities
$\bigstar \: \: \mathbf{\left(a+b\right)^{2} = a^{2}+b^{2}+2ab}$
$\bigstar \: \: \mathbf{\left(a-b\right)^{2} = a^{2}+b^{2}-2ab}$

$\implies \left(m+n\right)^{\frac{2}{3}}+ \left(m-n\right)^{\frac{2}{3}} = p^{\frac{2}{3}} \left[\left(sin^{2}\, \theta + cos^{2}\, \theta + 2sin\, \theta\: cos\, \theta\right) + \left(cos^{2}\, \theta + sin^{2}\, \theta - 2sin\, \theta\: cos\, \theta\right)\right]$



$\underline{\Large{\textsf{Step-8:}}}$
Using the Identity
$\bigstar\: \: \mathbf{sin^{2}\: \theta+ cos^{2}\: \theta = 1}$

$\implies \left(m+n\right)^{\frac{2}{3}}+ \left(m-n\right)^{\frac{2}{3}} = p^{\frac{2}{3}} \left[\left(1 + 2sin\, \theta\: cos\, \theta\right) + \left(1 - 2sin\, \theta\: cos\, \theta\right)\right]$

$\implies \left(m+n\right)^{\frac{2}{3}}+ \left(m-n\right)^{\frac{2}{3}} = p^{\frac{2}{3}} \left[2+ \cancel{2sin\, \theta\: cos\, \theta} - \cancel{2sin\, \theta\: cos\, \theta}\right]$

$\therefore\: \: \left(m+n\right)^{\frac{2}{3}}+ \left(m-n\right)^{\frac{2}{3}} = 2p^{\frac{2}{3}}$



$\mathbf{Hence\: Proved}$