Question

If x= (√a+√b)/(√a-√b) and y=(√a-√b)/(√a+√b) then x2+xy+y2=

Answer 1

x=(√a+√b)/(√a-√b) and y=(√a-√b)/(√a+√b)
∴, x+y=(√a+√b)/(√a-√b)+(√a-√b)/(√a+√b)
={(√a+√b)²+(√a-√b)²}/{(√a)²-(√b)²}
=(a+2√ab+b+a-2√ab+b)/(a-b)
=2(a+b)/(a-b) and
xy=(√a+√b)/(√a-√b)×(√a-√b)/(√a+√b)
=1
∴, x²+xy+y²
={(x+y)²-2xy}+xy
={2(a+b)/(a-b)}²-2×1+1
=4(a+b)²/(a-b)²-2+1
=4(a²+2ab+b²)/(a²-2ab+b²)-1
=(4a²+8ab+4b²-a²+2ab-b²)/(a²-2ab+b²)
=3a²+10ab+3b²/a²-2ab+b²

Answer 2

We use rationalization of denominators.

x y = 1  as  we can see that  y = 1/x

x = (√a + √b) / (√a - √b)
   = (√a + √b)² / [ (√a - √b) (√a + √b) ]
   = [a + b + 2 √(ab) ] / [a - b]

y = (√a - √b) / (√a + √b)
   = (√a - √b)² / [ (√a + √b) (√a - √b) ]
   = (a + b - 2 √(ab) ] / (a - b)

x+y = 2(a+b)/(a-b)
xy = 1

x² + xy + y²
  = (x + y)² - x y 
  = 4 (a+b)² / (a-b)²  - 1
  = [4 (a+b)²  - (a-b)² ] / (a-b)²
  = [ 3 a² + 3 b² + 10 ab ] / (a-b)²
    = 3 + 16 ab/(a-b)²