Question

a=√3+1/√3-1 and b=√3-1/√3+1 then the value of a2+ab+b2/a2-ab+b2 =?

Answer 1

${ \bold{ \underline{Given} : - }} \\ \\ a = \frac{ \sqrt{3} + 1}{ \sqrt{3} - 1 } \: \: and \: \: b = \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 } \\ \\ \\ { \bold{ \underline{ To \: \: find } : - }} \\ \\ \: \: \: \: \: . \: \: \: Value \: \: of \: \: {a}^{2} + ab + {b}^{2} \\ \: \: \: \: \: . \: \: \: Value \: \: of \: \: {a}^{2} - ab + {b}^{2} \\ \\ { \bold{ \underline{solution} : - }} \\ \\ = > First \: \: we \: \: have \: \: to \: \: find \: \: the \: \: values \: \: of \: \: {a}^{2} \: \: and \: \: {b}^{2} \\ \\ = > a = \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1} \\ \\ = > {a}^{2} = \frac{ {( \sqrt{3} + 1) }^{2} }{ {( \sqrt{3} - 1)}^{2} } \\ \\ = > { a}^{2} = \frac{3 + 1 + 2 \sqrt{3} }{3 + 1 - 2 \sqrt{3} } \\ \\ = > {a}^{2} = \frac{4 + 2 \sqrt{3} }{4 - 2 \sqrt{3} } = \frac{2 + \sqrt{3} }{2 - \sqrt{3} } \\ \\ = > {a}^{2} = \frac{2 + \sqrt{3} }{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} } \\ \\ = > {a}^{2} = \frac{ {(2 + \sqrt{ 3}) }^{2} }{ {(2)}^{2} - {( \sqrt{3} )}^{2} } \\ \\ = > {a}^{2} = 4 + 3 + 4 \sqrt{3} = 7 + 4 \sqrt{3} \\ \\ \: \: . \: \: now \: - \\ \\ = > b = \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 } \\ \\ = > {b}^{2} = \frac{ {( \sqrt{3} - 1)}^{2} }{ {( \sqrt{3} + 1) }^{2} } \\ \\ = > {b}^{2} = \frac{3 + 1 - 2 \sqrt{3} }{3 + 1 + 2 \sqrt{3} } \\ \\ = > {b}^{2} = \frac{4 - 2 \sqrt{3} }{4 + 2 \sqrt{3} } = \frac{2 - \sqrt{3} }{ 2+ \sqrt{3} } \\ \\ = > {b}^{2} = \frac{2 - \sqrt{3} }{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ = > {b}^{2} = \frac{ {(2 - \sqrt{3}) }^{2} }{ {(2)}^{2} - {( \sqrt{3}) }^{2} } \\ \\ = > {b}^{2} = 4 + 3 - 4 \sqrt{3} = 7 - 4 \sqrt{3} \\ \\ \\ \: \: \: \: . \: \: \:{ \bold{ ab = 1 }}\\ \\ \\ \: \: \: \: . \: \: so \: \: that - \\ \\ (1) \: \: {a}^{2} + ab + {b}^{2} = 7 + 4 \sqrt{3} + 1 + 7 - 4 \sqrt{3} \\ \\ \: \: \: \: \: \: \: \: \: = 14 + 1 = 15\\ \\ (2) \: \: \: {a}^{2} - ab + {b}^{2} = 7 + 4 \sqrt{3} - 1 + 7 - 4 \sqrt{3} \\ \\ \: \: \: \: \: \: \: \: \: \: = 14 - 1 = 13$