Math, asked by PagalLadiki143, 9 months ago

<marquee>please solve this

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Answered by RvChaudharY50
124

 \large{\mathfrak{\overbrace{\underbrace{\pink{\fbox{\green{\blue{\bf\:S}\pink{o}\red{l}\orange{u}\purple{t}\blue{i}\green{o}\red{n}}}}}}}}

\red\longmapsto\dfrac{2}{\sqrt[3]{9}-\sqrt[3]{3} + 1} - \dfrac{1}{\sqrt[3]{9} + \sqrt[3]{3} + 1}

\red\longmapsto\dfrac{2}{(\sqrt[3]{9} + 1)-\sqrt[3]{3}} - \dfrac{1}{(\sqrt[3]{9} +1) +  \sqrt[3]{3}}

 \blue{\textbf{Taking LCM of Denominator and using (a + b)(a - b) = (a² - b²) Now,}}

\red\longmapsto\dfrac{2( \sqrt[3]{9} +  \sqrt[3]{3} + 1) - 1( \sqrt[3]{9} -  \sqrt[3]{3} + 1) }{( \sqrt[3]{9} + 1)^{2} -  (\sqrt[3]{3})^{2}  }

  \green{\bf \: using \: \sqrt[a]{b} \: as \:  {b}^{ \frac{1}{a}} \: Now,}

\red\longmapsto\dfrac{2 \times  {9}^{\frac{1}{3}} + 2 \times {3}^{ \frac{1}{3}} + 2 -  {9}^{ \frac{1}{3}} +  {3}^{ \frac{1}{3}} - 1}{(1 +  {9}^{ \frac{1}{3}}) ^{2} -(3^{ \frac{1}{3}})^{2}}

 \pink{\textsf{Now using (a + b)² = a² + b² + 2ab in Denominator ,}}

\red\longmapsto\dfrac{9^{\frac{1}{3}} + 3 \times  {3}^{ \frac{1}{3}} + 1}{1 +  {9}^{ \frac{2}{3}} + 2 \times  {9}^{ \frac{1}{3}} -  {3}^{ \frac{2}{3}}}

\red\longmapsto \:  \dfrac{ {9}^{ \frac{1}{3}} +  {3}^{ \frac{4}{3}} + 1}{1 +  {9}^{ \frac{2}{3}} +  {3}^{\frac{2}{3}}}

\red\longmapsto \:  \dfrac{ {3}^{ \frac{2}{3} } +  {3}^{ \frac{4}{3} } + 1 }{{3}^{ \frac{2}{3} } +  {3}^{ \frac{4}{3} } + 1}

\red\longmapsto\:\rm \: \bold{\boxed{\large{\boxed{\orange{\small{\boxed{\large{\red{\bold{1}}}}}}}}}}

\rule{200}{4}

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