Question

if a=root 3 +1/root 3-1 and b = root 3 - 1/root 3 +1, then what is the value of a²+ab+b²​

Answer 1

Answer:

this is answers and it's right

Answer 2

HERE ,

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

TO FIND : the value of a²+ab+b²

~~

${a}^{2} = ( { \frac{ \sqrt{3} + 1}{ \sqrt{3} - 1 } })^{2} \\ = \frac{ { \sqrt{3} }^{2} + 2 \sqrt{3} \times 1 + {1}^{2} }{ { \sqrt{3} }^{2} - 2 \sqrt{3} \times 1 + {1}^{2} } \\ = \frac{3 + 2 \sqrt{3} + 1}{3 - 2 \sqrt{3} + 1} \\ = \frac{4 + 2 \sqrt{3} }{2 - 2 \sqrt{3} } \\ = \frac{2 + \sqrt{3} }{1 - \sqrt{3} }$

$ab = \frac{( \sqrt{3} - 1)}{( \sqrt{3} + 1) } \times \frac{( \sqrt{3} + 1)}{ (\sqrt{3} - 1) } \\ = 1$

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

SO,

a²+ab+b²

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