Question

If x=√3+1÷√3-1,y=√3-1÷√3+1 then find the value of x2+y2+xy​

Answer 1

Answer:

x6+y6+xy(square)

Step-by-step explanation:

wait two mins i write on paper

Answer 2

Answer:

15

Step-by-step explanation:

We have,

x = (√3+1) ÷  (√3-1)

y =  (√3-1) ÷ (√3+1)

TO Find - x^{2} + y^{2} + xy

= {(√3+1) ÷  (√3-1)}^{2} + (√3 - 1) ÷ (√3 + 1) + { (√3+1) ÷  (√3-1)} × {(√3-1) ÷ (√3+1)}

Formula

(a + b)^{2} = a^{2} + b^{2} + 2ab

So, x^{2} + y^{2} + xy = {(√3+1) ÷  (√3-1)}^{2} + (√3 - 1) ÷ (√3 + 1) + { (√3+1) ÷  (√3-1)} × {(√3-1) ÷ (√3+1)}

= {(3 + 1 + 2√3) ÷ (3 + 1 - 2√3)} +  {(3 + 1 - 2√3) ÷ (3 + 1 + 2√3)} + 1

= {(4 + 2√3) ÷ (4 - 2√3)} + {(4 - 2√3) ÷ (4 + 2√3)} + 1

= {(2 + √3) ÷ (2 - √3)} + {(2 - √3) ÷ (2 + √3)} + 1  

= [(2 + √3)^{2} + (2 - √3)^{2} + (2 + √3)(2 - √3)] ÷ [(2 + √3)(2 - √3)]

= [(4 + 3 + 4√3) + (4 + 3 - 4√3) + ( 4 - 3)] ÷ (4 - 3)

= [ (7 + 4√3) + (7 - 4√3) + 1 ]  ÷ ( 1 )

= ( 14 + 1 )

= 15

Hence    x^{2} + y^{2} + xy = 15