If `a = (sqrt5 + 1)/(sqrt5 + 1) and b = (sqrt5 -1)/(sqrt5 + 1)`, then find the value of
(a) `(a^(2) + ab + b^(2))/(a^(2) - ab + b^(2))` (b) `((a -b)^(3))/((a + b)^(3))`
(c) `(3a^(2) + 5ab + b^(2))/(3a^(2) - 5ab + b^(2))` (d) `(a^(3) + b^(3))/(a^(3) - b^(3))`

Explanation

Given that `a = (sqrt5 + 1)/( sqrt5 -1) and b = (sqrt5 -1)/(sqrt5 + 1)`
`:. a - b = (sqrt5 + 1)/(sqrt5 -1) - (sqrt5 -1)/(sqrt5 + 1)`
`= ((sqrt5 + 1)^(2) - (sqrt5 -1)^(2))/((sqrt5-1) (sqrt5+1)) = (4.sqrt5.1)/((sqrt5)^(2) - (1)^(2)) = (4 sqrt5)/(5 -1)`
`= (4 sqrt5)/(4) = sqrt5`
`a + b = (sqrt5 + 1)/(sqrt5 -1) + (sqrt5 -1)/(sqrt5 + 1)`
`= ((sqrt5 + 1)^(2) + (sqrt5 -1)^(2))/((sqrt5 -1) (sqrt5 + 1)) = (2{(sqrt5)^(2) + (1)^(2)})/((sqrt5)^(2) - (1)^(2)) = (2 (5 + 1))/(5 -1)`
`= (12)/(4) = 3`
and `ab = (sqrt5 + 1)/(sqrt5 -1) xx (sqrt5 -1)/(sqrt5 + 1) = 1`
Now, (a) `(a^(2) + ab + b^(2))/(a^(2) - ab + b^(2)) = (a^(2) + 2ab + b^(2) - ab)/(a^(2) - 2ab + b^(2) + ab)`
`= ((a + b)^(2) - ab)/((a -b)^(2) + ab) = ((3)^(2) -1)/((sqrt5)^(2) + 1) = (9 -1)/(5 + 1) = (8)/(6) = (4)/(3) = 1 (1)/(3)`
(b) `((a - b)^(3))/((a + b)^(3)) = ((sqrt5)^(3))/((3)^(3)) = (5 sqrt5)/(27)`
(c) `(3a^(2) + 5ab + 3b^(2))/(3a^(2) - 5ab + 3b^(2)) = (3a^(2) + 6ab + 3b^(2) - ab)/(3a^(2) - 6ab + 3b^(2) + ab)`
`= (3 (a^(2) + 2ab + b^(2)) - ab)/(3(a^(2) - 2ab + b^(2)) + ab) = (3 (a + b)^(2)- ab)/(3 (a -b)^(2) + ab) = (3 (3)^(2) -1)/(3 (sqrt5)^(2) + 1)`
`= (3 xx 9 -1)/(3 xx 5 + 1) = (27 -1)/(15 + 1) = (26)/(16) = (13)/(8) = 1 (5)/(8)`
(d) `(a^(3) + b^(3))/(a^(3) - b^(3)) = ((a + b)^(3) - 3ab (a + b))/((a -b)^(3) + 3ab (a -b))`
`= ((3)^(3) - 3.1.3)/((sqrt5)^(3) + 3.1.sqrt5) = (27 - 9)/(5 sqrt5 + 3 sqrt5) = (18)/(8sqrt5) = (9)/(4 sqrt5) = (9 sqrt5)/(20)`