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Question Number 78857 by mr W last updated on 21/Jan/20

w i t h a, b \in R p r o v e t h a t

\sqrt[3]{a + \sqrt{3} b i} + \sqrt[3]{a - \sqrt{3} b i}

h a s a l w a y s r e a l v a l u e a n d f i n d t h i s

v a l u e (o r a w a y h o w t o f i n d) .

e x a m p l e s :

a = 15, b = \frac{28}{9} \Rightarrow r e s u l t = 5

a = 6, b = \frac{35}{9} \Rightarrow r e s u l t = 4

a = - 24, b = \frac{80}{9} \Rightarrow r e s u l t = 4

Commented by mr W last updated on 21/Jan/20

i^{'} m n o t s u r e i f t h e s t a t e m e n t i s c o r r e c t .

d i s c u s s i o n i s w e l c o m e!

Commented by john santu last updated on 22/Jan/20

l e t u = \sqrt[3]{a + b i \sqrt{3}} \Rightarrow u^{3} = a + b i \sqrt{3}

v = \sqrt[3]{a - b i \sqrt{3}} \Rightarrow v^{3} = a - b i \sqrt{3}

u + v = \frac{u^{3} + v^{3}}{u^{2} - u v + v^{2}}

u^{3} + v^{3} = 2 a

u^{2} - u v + v^{2} = \sqrt[3]{a^{2} - 3 b^{2} + 2 a b i \sqrt{3}} -

\sqrt[3]{a^{2} + 3 b^{2}} + \sqrt[3]{a^{2} + 3 b^{2} - 2 a b i \sqrt{3}}

Commented by john santu last updated on 21/Jan/20

d i f f i c u l t t o w r i t e s i r . t o o l o n g

Commented by mr W last updated on 21/Jan/20

t h a n k s . b u t n o t y e t u n d e r s t o o d w h a t

y o u m e a n .

Commented by john santu last updated on 21/Jan/20

w a i t s i r

Answered by MJS last updated on 21/Jan/20

r = α^{\frac{1}{3}} + β^{\frac{1}{3}}

r^{3} - 3 α^{\frac{1}{3}} β^{\frac{1}{3}} r - (α + β) = 0

α = p + q \land β = p - q

r^{3} - 3 {(p^{2} - q^{2})}^{\frac{1}{3}} r - 2 p = 0

p = a \land q = \sqrt{3} b i

r^{3} - 3 {(a^{2} + 3 b^{2})}^{\frac{1}{3}} r - 2 a = 0

D = - 3 b^{2} ⩽ 0 \forall b \in R \Rightarrow 3 real solutions

but obviously

\sqrt[n]{s + t i} = u + v i \land \sqrt[n]{s - t i} = u - v i \Rightarrow

\sqrt[n]{s + t i} + \sqrt[n]{s - t i} = 2 u \in R

Commented by MJS last updated on 21/Jan/20

\dots we can try factors of \pm 2 a but usually we

need the trigonometric solution and obviously

we {don}^{'} t get “ {nice}^{″} solutions in most cases

your examples lead to {(a^{2} + 3 b^{2})}^{\frac{1}{3}} = c \in Z, we

might get some “ {nice}^{″} solutions for c \in Q but

not for c \in R ∖ Q

Commented by mr W last updated on 21/Jan/20

t h a n k y o u s i r! v e r y i n f o r m a t i v e a n s w e r!

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