Question

$\ {if \: \frac{1}{ \sqrt{5} + 2 } + \frac{1}{2 + \sqrt{3} } + \frac{1}{ \sqrt{3} + \sqrt{2} } + \frac{1}{ \sqrt{2} + 1 }} \\ \\ \tt{ = a + b \sqrt{c} \: \: then \: find \: the \: value \: of \: {a}^{2} + {b}^{2} + ab }$

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Answer 1

${\underline{\large{\pmb{\frak{Given...}}}}}$

$\bullet \; {\underline{\boxed{\tt{ \frac{1}{\sqrt{5} + 2 } + \frac{1}{2 + \sqrt{3} } + \frac{1}{\sqrt{3} + \sqrt{2} } + \frac{1}{\sqrt{2} + 1 } = a + b\sqrt{c} }}}}$

${\underline{\large{\pmb{\frak{To \; Find...}}}}}$

★ The value of the expression a² + b² + ab

${\underline{\large{\pmb{\frak{Full \; solution...}}}}}$

★ Now firstly let's rewrite the equation given and rationalise the denominators

${: \implies } \bf \dfrac{1}{\sqrt{5} + 2 } + \dfrac{1}{2 + \sqrt{3} } + \dfrac{1}{\sqrt{3} + \sqrt{2} } + \dfrac{1}{\sqrt{2} + 1 } = a + b\sqrt{c}$

• Rationalising the denominator

${ : \implies } \bf \bigg[\dfrac{1}{\sqrt{5} + 2 } \times \dfrac{\sqrt{5} - 2 }{\sqrt{5} - 2} \bigg]+ \bigg[\dfrac{1}{2 + \sqrt{3} } \times \dfrac{ 2 - \sqrt{3} }{2 - \sqrt{3} } \bigg] + \bigg[ \dfrac{1}{\sqrt{3} + \sqrt{2} } \times \dfrac{\sqrt{3} - \sqrt{2} }{\sqrt{3} - \sqrt{2}} \bigg] + \bigg[\dfrac{1}{\sqrt{2} +1} \times \dfrac{\sqrt{2} - 1 }{\sqrt{2} -1} \bigg]$

• Using suitable algebric Identity

$\bigstar \; {\underline{\boxed{\bf{ ( A + B ) ( A - B ) = A^2 - B^2 }}}}$

• Solving the equation in further

${ : \implies } \bf \bigg[ \dfrac{\sqrt{5} - 2 }{5 - 4} \bigg]+ \bigg[\dfrac{2 - \sqrt{3} } {4 - 3 } \bigg] + \bigg[ \dfrac{\sqrt{3} - \sqrt{2} }{ 3 - 2} \bigg] + \bigg[\dfrac{\sqrt{2} - 1}{ 2- 1} \bigg] = a + b \sqrt{c} \\ \\ \\ { : \implies } \bf \bigg[ \dfrac{\sqrt{5} - 2 }{1} \bigg]+ \bigg[\dfrac{2 - \sqrt{3} } {1 } \bigg] + \bigg[ \dfrac{\sqrt{3} - \sqrt{2} }{ 1} \bigg] + \bigg[\dfrac{\sqrt{2} - 1}{ 1} \bigg]= a + b \sqrt{c}$

• Canceling the denominator since it's 1

${: \implies } \bf \sqrt{5} - \cancel 2 + \cancel 2 - \cancel {\sqrt{3} }+ \cancel{\sqrt{3}} - \cancel {\sqrt{2} }+ \cancel{\sqrt{2}} = a + b \sqrt{c} \\ \\ \\ {: \implies } \bf \sqrt{5} - 1 = a + b\sqrt{c} \\ \\ \\ {: \implies } \bf - 1 + 1 \sqrt{5} = a + b \sqrt{c}$

• Finding the value of the required expression

${: \implies } \bf a^2 + b ^2 + ab = ( - 1 )^2 + 1^2 + ( - 1) ( 1 ) \\\\\\ {: \implies } \bf a^2 + b^2 + ab = 1 + 1 - 1 \\ \\ \\{: \implies } {\underline{\boxed{\bf{ a^2 + b ^ 2 + ab = 1 }}}\bigstar}$

• Henceforth the value of a² + b² + ab = 1

${\underline{\large{\pmb{\frak{Additional \; Information...}}}}$

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c} \\ \tiny\bf{\dag}\:\underline{\frak{\rm{S}\frak{ome\:important\:algebric\:identities\:::}}} \\\\ \green{\bigstar}\:\rm \red{ (A+B)^{2} = A^{2} + 2AB + B^{2}} \\\\ \red{\bigstar}\rm\: \green{(A-B)^{2} = A^{2} - 2AB + B^{2}} \\\\ \orange{\bigstar}\rm\: \blue{A^{2} - B^{2} = (A+B)(A-B)}\\\\ \blue{\bigstar}\rm\: \orange{(A+B)^{2} = (A-B)^{2} + 4AB}\\\\ \pink{\bigstar}\rm\: \purple{(A-B)^{2} = (A+B)^{2} - 4AB}\\\\ \purple{\bigstar} \rm\: \pink{(A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}}\\\\ \gray{\bigstar}\rm\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\ \bigstar\rm\: \gray{A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})} \\\\ \end{array}}\end{gathered}\end{gathered}$

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