Tính: `\root{3}{9 + 4\sqrt{5}} + \root{3}{9 - 4\sqrt{5}}`

Đặt: `\root{3}{9 + 4\sqrt{5}} = a`

`\root{3}{9 - 4\sqrt{5}} = b`

Ta có:

`a^3 + b^3 = 9 + 4\sqrt{5} + (9 - 4\sqrt{5}) = 18`

`ab = \root{3}{(9 + 4\sqrt{5})(9 - 4\sqrt{5})} = 1`

Ta có:

`(a + b)^3`

`= a^3 + b^3 + 3ab(a + b)`

`= 18 + 3.1.(a + b)`

`⇔ (a + b)^3 - 3(a + b) - 18 = 0`

`⇔ (a + b)^3 - 3(a + b)^2 + 3(a + b)^2 - 9(a + b) + 6(a + b) - 18 = 0`

`⇔ (a + b)^2(a + b - 3) + 3(a + b)(a + b - 3) + 6(a + b - 3) = 0`

`⇔ (a + b - 3)[(a + b)^2 + 3(a + b) + 6] = 0`

Vì: `(a + b)^2 + 3(a + b) + 6 = (a + b + 3/2)^2 + 15/4 > 0`

`⇒ a + b - 3 = 0`

`⇔ a + b = 3`

Vậy: `\root{3}{9 + 4\sqrt{5}} + \root{3}{9 - 4\sqrt{5}} = 3`