rationalize-the-denominator-1-a-b-1-3-c-1-3-

Question Number 210324 by Ghisom last updated on 06/Aug/24

rationalize the denominator :

\frac{1}{a + b^{1 / 3} + c^{1 / 3}}

Commented by Spillover last updated on 06/Aug/24

a n s

\frac{a^{2} - {a b}^{\frac{1}{3}} + b^{\frac{2}{3}} - {a c}^{\frac{1}{3}} - b^{\frac{1}{3}} c^{\frac{1}{3}} + c^{\frac{2}{3}}}{a^{3} + b + c - 3 {a b}^{\frac{1}{3}} c^{\frac{1}{3}}}

Commented by Ghisom last updated on 06/Aug/24

{there}^{'} s still - 3 a {(b c)}^{1 / 3}

Answered by Frix last updated on 08/Aug/24

1 . Expand with [ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i]

(a + ω b^{\frac{1}{3}} + ω^{2} c^{\frac{1}{3}}) (a + ω^{2} b^{\frac{1}{3}} + ω c^{\frac{1}{3}}) =

= (a^{2} - a (b^{\frac{1}{3}} + c^{\frac{1}{3}}) + b^{\frac{2}{3}} - b^{\frac{1}{3}} c^{\frac{1}{3}} + c^{\frac{2}{3}})

The denominator now is

(a^{3} + b + c - 3 {a b}^{\frac{1}{3}} c^{\frac{1}{3}}) = (p - q^{\frac{1}{3}})

\Rightarrow

2 . Expand with [p = a^{3} + b + c \land q = 27 a^{3} b c]

(p^{2} + {p q}^{\frac{1}{3}} + q^{\frac{2}{3}}) to get p^{3} - q

The denominator now is

a^{9} + 3 a^{6} (b + c) + 3 a^{3} (b^{2} - 7 b c + c^{2}) + {(b + c)}^{3}

[Works even with a = α^{\frac{1}{3}}]

Commented by Ghisom last updated on 14/Aug/24

thank you

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