Question

if a = 1/6 - root 35, and b =1/6 +root 35 then find a^2b+ ab^2

Answer 1

Answer:

$a = \frac{1}{6 - \sqrt{35} } = \frac{6 + \sqrt{35} }{36 - 35} = 6 + \sqrt{35} \\ b = \frac{1}{6 + \sqrt{35} } = \frac{6 - \sqrt{35} }{36 - 35} = 6 - \sqrt{35} \\ {a}^{2} b + a {b}^{2} = ab(a + b) \\ = (6 + \sqrt{35} )(6 - \sqrt{35} )(6 + \sqrt{35} + 6 - \sqrt{35} ) \\ = (36 - 35)(12) \\ = (1)(12) \\ = 12$