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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

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Question Number 78857 by mr W last updated on 21/Jan/20
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
$${with}\:{a},{b}\in{R}\:{prove}\:{that} \\ $$$$\sqrt[{\mathrm{3}}]{{a}+\sqrt{\mathrm{3}}{bi}}+\sqrt[{\mathrm{3}}]{{a}−\sqrt{\mathrm{3}}{bi}} \\ $$$${has}\:{always}\:{real}\:{value}\:{and}\:{find}\:{this} \\ $$$${value}\:\left({or}\:{a}\:{way}\:{how}\:{to}\:{find}\right). \\ $$$${examples}: \\ $$$${a}=\mathrm{15},\:{b}=\frac{\mathrm{28}}{\mathrm{9}}\:\:\Rightarrow\:{result}=\mathrm{5} \\ $$$${a}=\mathrm{6},\:{b}=\frac{\mathrm{35}}{\mathrm{9}}\:\:\Rightarrow\:{result}=\mathrm{4} \\ $$$${a}=−\mathrm{24},\:{b}=\frac{\mathrm{80}}{\mathrm{9}}\:\:\Rightarrow\:{result}=\mathrm{4} \\ $$
Commented by mr W last updated on 21/Jan/20
i′m not sure if the statement is correct. discussion is welcome!
$${i}'{m}\:{not}\:{sure}\:{if}\:{the}\:{statement}\:{is}\:{correct}. \\ $$$${discussion}\:{is}\:{welcome}! \\ $$
Commented by john santu last updated on 22/Jan/20
let u= ((a+bi(√3) ))^(1/3) ⇒u^3 =a+bi(√3) v =((a−bi(√3)))^(1/3) ⇒v^3 =a−bi(√3) u+v = ((u^3 +v^3 )/(u^2 −uv+v^2 )) u^3 +v^3 = 2a u^2 −uv+v^2 = ((a^2 −3b^2 +2abi(√3)))^(1/3) − ((a^2 +3b^2 ))^(1/3) +((a^2 +3b^2 −2abi(√3)))^(1/3)
$${let}\:{u}=\:\sqrt[{\mathrm{3}}]{{a}+{bi}\sqrt{\mathrm{3}}\:}\:\Rightarrow{u}^{\mathrm{3}} ={a}+{bi}\sqrt{\mathrm{3}} \\ $$$${v}\:=\sqrt[{\mathrm{3}}]{{a}−{bi}\sqrt{\mathrm{3}}}\:\Rightarrow{v}^{\mathrm{3}} ={a}−{bi}\sqrt{\mathrm{3}} \\ $$$${u}+{v}\:=\:\frac{{u}^{\mathrm{3}} +{v}^{\mathrm{3}} }{{u}^{\mathrm{2}} −{uv}+{v}^{\mathrm{2}} } \\ $$$${u}^{\mathrm{3}} +{v}^{\mathrm{3}} \:=\:\mathrm{2}{a}\: \\ $$$${u}^{\mathrm{2}} −{uv}+{v}^{\mathrm{2}} =\:\sqrt[{\mathrm{3}}]{{a}^{\mathrm{2}} −\mathrm{3}{b}^{\mathrm{2}} +\mathrm{2}{abi}\sqrt{\mathrm{3}}}\:− \\ $$$$\sqrt[{\mathrm{3}}]{{a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} }\:+\sqrt[{\mathrm{3}}]{{a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} −\mathrm{2}{abi}\sqrt{\mathrm{3}}} \\ $$
Commented by john santu last updated on 21/Jan/20
difficult to write sir. too long
$${difficult}\:{to}\:{write}\:{sir}.\:{too}\:{long} \\ $$
Commented by mr W last updated on 21/Jan/20
thanks. but not yet understood what you mean.
$${thanks}.\:{but}\:{not}\:{yet}\:{understood}\:{what}\: \\ $$$${you}\:{mean}. \\ $$
Commented by john santu last updated on 21/Jan/20
wait sir
$${wait}\:{sir} \\ $$
Answered by MJS last updated on 21/Jan/20
r=α^(1/3) +β^(1/3) r^3 −3α^(1/3) β^(1/3) r−(α+β)=0 α=p+q∧β=p−q r^3 −3(p^2 −q^2 )^(1/3) r−2p=0 p=a∧q=(√3)bi r^3 −3(a^2 +3b^2 )^(1/3) r−2a=0 D=−3b^2 ≤0∀b∈R ⇒ 3 real solutions but obviously ((s+ti))^(1/n) =u+vi ∧ ((s−ti))^(1/n) =u−vi ⇒ ((s+ti))^(1/n) +((s−ti))^(1/n) =2u ∈R
$${r}=\alpha^{\frac{\mathrm{1}}{\mathrm{3}}} +\beta^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${r}^{\mathrm{3}} −\mathrm{3}\alpha^{\frac{\mathrm{1}}{\mathrm{3}}} \beta^{\frac{\mathrm{1}}{\mathrm{3}}} {r}−\left(\alpha+\beta\right)=\mathrm{0} \\ $$$$\alpha={p}+{q}\wedge\beta={p}−{q} \\ $$$${r}^{\mathrm{3}} −\mathrm{3}\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} {r}−\mathrm{2}{p}=\mathrm{0} \\ $$$${p}={a}\wedge{q}=\sqrt{\mathrm{3}}{b}\mathrm{i} \\ $$$${r}^{\mathrm{3}} −\mathrm{3}\left({a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} {r}−\mathrm{2}{a}=\mathrm{0} \\ $$$${D}=−\mathrm{3}{b}^{\mathrm{2}} \leqslant\mathrm{0}\forall{b}\in\mathbb{R}\:\Rightarrow\:\mathrm{3}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$ \\ $$$$\mathrm{but}\:\mathrm{obviously} \\ $$$$\sqrt[{{n}}]{{s}+{t}\mathrm{i}}={u}+{v}\mathrm{i}\:\wedge\:\sqrt[{{n}}]{{s}−{t}\mathrm{i}}={u}−{v}\mathrm{i}\:\Rightarrow \\ $$$$\sqrt[{{n}}]{{s}+{t}\mathrm{i}}+\sqrt[{{n}}]{{s}−{t}\mathrm{i}}=\mathrm{2}{u}\:\in\mathbb{R} \\ $$
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
$$…\mathrm{we}\:\mathrm{can}\:\mathrm{try}\:\mathrm{factors}\:\mathrm{of}\:\pm\mathrm{2}{a}\:\mathrm{but}\:\mathrm{usually}\:\mathrm{we} \\ $$$$\mathrm{need}\:\mathrm{the}\:\mathrm{trigonometric}\:\mathrm{solution}\:\mathrm{and}\:\mathrm{obviously} \\ $$$$\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{get}\:“\mathrm{nice}''\:\mathrm{solutions}\:\mathrm{in}\:\mathrm{most}\:\mathrm{cases} \\ $$$$\mathrm{your}\:\mathrm{examples}\:\mathrm{lead}\:\mathrm{to}\:\left({a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} ={c}\in\mathbb{Z},\:\mathrm{we} \\ $$$$\mathrm{might}\:\mathrm{get}\:\mathrm{some}\:“\mathrm{nice}''\:\mathrm{solutions}\:\mathrm{for}\:{c}\in\mathbb{Q}\:\mathrm{but} \\ $$$$\mathrm{not}\:\mathrm{for}\:{c}\in\mathbb{R}\backslash\mathbb{Q} \\ $$
Commented by mr W last updated on 21/Jan/20
thank you sir! very informative answer!
$${thank}\:{you}\:{sir}!\:{very}\:{informative}\:{answer}! \\ $$

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