Question

If a=√3+1/√3-1and b=√3-1/√3+1, find the value of a^2+ab+b^2/a^2-ab+b^2​

Answer 1

Given-

$a = \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1 } \: and \: \: b = \frac{ \sqrt{3} - 1}{ \sqrt{3} + 1 }$

To find -

$the \: value \: of \: \: \frac{ {a}^{2} + ab + {b}^{2} }{ {a}^{2} - ab + {b}^{2} }$

Solution -

$a = \frac{ \sqrt{3} + 1}{ \sqrt{3} - 1}$

By rationalising the denominator we will get -

$a = \frac{ {( \sqrt{3} + 1)}^{2} }{( \sqrt{3} - 1)( \sqrt{3} + 1) }$

$a = \frac{ {( \sqrt{3} )}^{2} + {(1)}^{2} + 2 \sqrt{3} }{ {( \sqrt{3} }^{2} ) - {(1)}^{2} }$

$a = \frac{3 + 1 + 2 \sqrt{3} }{3 - 1}$

$a = \frac{4 + 2 \sqrt{3} }{2}$

$a = 2 + \sqrt{3}$

And,

$b = \frac{ \sqrt{3} - 1}{ \sqrt{3} + 1}$

By rationalising the denominator we will get -

$b = \frac{ {( \sqrt{3} - 1)}^{2} }{( \sqrt{3} + 1)( \sqrt{3} - 1) }$

$b = \frac{ { (\sqrt{3}) {}^{2} + ({1})^{2} - 2 \sqrt{3} } }{ { \sqrt{3} }^{2} - {1}^{2} }$

$b = \frac{3 + 1 - 2 \sqrt{3} }{3 - 1}$

$b = \frac{4 - 2 \sqrt{3} }{2}$

$b = 2 - \sqrt{3}$

Now,

$= \frac{ {a}^{2} + ab + {b}^{2} }{ {a}^{2} - ab + {b}^{2} }$

By putting the values we will get -

$= \frac{ {(2 + \sqrt{3}) }^{2} + (2 + \sqrt{3} )(2 - \sqrt{3} ) + {(2 - \sqrt{3} )}^{2} }{ {(2 + \sqrt{3} )}^{2} - (2 + \sqrt{3} )(2 - \sqrt{3} ) + ( {2 - \sqrt{3} )}^{2} }$

$= \frac{ {(2)}^{2} + {( \sqrt{3}) }^{2} + 4 \sqrt{3} + {(2)}^{2} - {( \sqrt{3}) }^{2} + {(2)}^{2} + {( \sqrt{3}) }^{2} - 4 \sqrt{3} }{ {(2)}^{2} + {( \sqrt{3}) }^{2} + 4 \sqrt{3} - {(2)}^{2} - {( \sqrt{3}) }^{2} + {(2)}^{2} + {( \sqrt{3}) }^{2} - 4 \sqrt{3} }$

$= \frac{(4 + 3 + 4 \sqrt{3}) + (4 - 3) +( 4 + 3 - 4 \sqrt{3} )}{(4 + 3 + 4 \sqrt{3} )- (4 - 3) + (4 + 3 - 4 \sqrt{3}) }$

$= \frac{7 + 4 \sqrt{3} + 1 + 7 - 4 \sqrt{3} }{7 + 4 \sqrt{3} - 1 + 7 - 4 \sqrt{3} }$

$= \frac{15}{13}$

Therefore

$the \: value \: of \: \frac{ {a}^{2} + ab + {b}^{2} }{ {a}^{2} - ab + {b}^{2} } \: is \: \: \frac{15}{13}$