with a,b∈R prove that ((a+(√3)bi))^(1/3) +((a−(√3)bi))^(1/3) has always real value and find this value (or a way how to find). examples: a=15, b=((28)/9) ⇒ result=5 a=6, b=((35)/9) ⇒ result=4 a=−24, b=((80)/9) ⇒ result=4


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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
	$...we can try factors of ±2a but usually we need the trigonometric solution and obviously we don′t get “nice” solutions in most cases your examples lead to (a^2 +3b^2 )^(1/3) =c∈Z, we might get some “nice” solutions for c∈Q but not for c∈R\Q$
	$. . . we can try factors of \pm 2 a but usually we$ $need the trigonometric solution and obviously$ $we {don}^{'} t get ‘ ‘ {nice}^{″} solutions in most cases$ $your examples lead to {(a^{2} + 3 b^{2})}^{\frac{1}{3}} = c \in Z, we$ $might get some ‘ ‘ {nice}^{″} solutions for c \in Q but$ $not for c \in R ∖ Q$
	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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