F3 maths 關於恆等式

(1)(y + 1/y)^2 = y^2 + 2 + 1/y^2.
But y^2 + 1/y^2 = 7. That is (y+1/y)^2 = 7 + 2 = 9. Therefore, y+1/y = sqrt9 = 3.
(2)(y-1/y)^2 = y^2 + 1/y^2 - 2 = 7 - 2 = 5.
(3)
y^2 - 1/y^2 = (y-1/y)(y+1/y) = sqrt5 x 3 = 3sqrt5.
Therefore, (y^2 - 1/y^2)^2 = (3sqrt5)^2 = 9 x 5 = 45.