Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 140774 by ajfour last updated on 12/May/21

Prove that ((1/( (√2)))+(5/(3(√6))))^(1/3) +((1/( (√2)))−(5/(3(√6))))^(1/3) = (√2)

$${Prove}\:{that} \\ $$$$\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{5}}{\mathrm{3}\sqrt{\mathrm{6}}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{5}}{\mathrm{3}\sqrt{\mathrm{6}}}\right)^{\mathrm{1}/\mathrm{3}} =\:\sqrt{\mathrm{2}} \\ $$

Commented by ajfour last updated on 12/May/21

All are Excellent solutions : cant we then always have a compact answer to (a+ib)^(1/3) +(a−ib)^(1/3) =x without trigo functions, that is.

$${All}\:{are}\:{Excellent}\:{solutions}\:: \\ $$$${cant}\:{we}\:{then}\:{always}\:{have}\:{a} \\ $$$${compact}\:{answer}\:{to} \\ $$$$\:\:\left({a}+{ib}\right)^{\mathrm{1}/\mathrm{3}} +\left({a}−{ib}\right)^{\mathrm{1}/\mathrm{3}} ={x} \\ $$$${without}\:{trigo}\:{functions},\:{that}\:{is}. \\ $$

Commented by MJS_new last updated on 13/May/21

p^(1/3) +q^(1/3) =x p+q+3p^(1/3) q^(1/3) (p^(1/3) +q^(1/3) )=x^3 3p^(1/3) q^(1/3) x=x^3 −p−q x^9 −3(p+q)x^6 +3(p^2 −7pq+q^2 )x^3 −(p+q)^3 =0 p=a+bi∧q=a−bi x^9 −6ax^6 −3(5a^2 +9b^2 )x^3 −8a^3 =0 x=t^(1/3) t^3 −6at^2 −3(5a^2 +9b^2 )t−8a^3 =0 t=z+2a z^3 −27(a^2 +b^2 )z−54a(a^2 +b^2 )=0 only if this has a “nice” solution...

$${p}^{\mathrm{1}/\mathrm{3}} +{q}^{\mathrm{1}/\mathrm{3}} ={x} \\ $$$${p}+{q}+\mathrm{3}{p}^{\mathrm{1}/\mathrm{3}} {q}^{\mathrm{1}/\mathrm{3}} \left({p}^{\mathrm{1}/\mathrm{3}} +{q}^{\mathrm{1}/\mathrm{3}} \right)={x}^{\mathrm{3}} \\ $$$$\mathrm{3}{p}^{\mathrm{1}/\mathrm{3}} {q}^{\mathrm{1}/\mathrm{3}} {x}={x}^{\mathrm{3}} −{p}−{q} \\ $$$${x}^{\mathrm{9}} −\mathrm{3}\left({p}+{q}\right){x}^{\mathrm{6}} +\mathrm{3}\left({p}^{\mathrm{2}} −\mathrm{7}{pq}+{q}^{\mathrm{2}} \right){x}^{\mathrm{3}} −\left({p}+{q}\right)^{\mathrm{3}} =\mathrm{0} \\ $$$${p}={a}+{b}\mathrm{i}\wedge{q}={a}−{b}\mathrm{i} \\ $$$${x}^{\mathrm{9}} −\mathrm{6}{ax}^{\mathrm{6}} −\mathrm{3}\left(\mathrm{5}{a}^{\mathrm{2}} +\mathrm{9}{b}^{\mathrm{2}} \right){x}^{\mathrm{3}} −\mathrm{8}{a}^{\mathrm{3}} =\mathrm{0} \\ $$$${x}={t}^{\mathrm{1}/\mathrm{3}} \\ $$$${t}^{\mathrm{3}} −\mathrm{6}{at}^{\mathrm{2}} −\mathrm{3}\left(\mathrm{5}{a}^{\mathrm{2}} +\mathrm{9}{b}^{\mathrm{2}} \right){t}−\mathrm{8}{a}^{\mathrm{3}} =\mathrm{0} \\ $$$${t}={z}+\mathrm{2}{a} \\ $$$${z}^{\mathrm{3}} −\mathrm{27}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){z}−\mathrm{54}{a}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\mathrm{only}\:\mathrm{if}\:\mathrm{this}\:\mathrm{has}\:\mathrm{a}\:``\mathrm{nice}''\:\mathrm{solution}... \\ $$

Commented by ajfour last updated on 13/May/21

Thanks Sir, i dont think it always be as nice..

Answered by liberty last updated on 12/May/21

a+b = ((a^3 +b^3 )/(a^2 −ab+b^2 )) ((1/( (√2)))+(5/(3(√6))))^(1/3) +((1/( (√2)))−(5/(3(√6))))^(1/3) = ((√2)/([((1/( (√2)))+(5/(3(√6))))^2 ]^(1/3) −((1/(27)))^(1/(3 )) + [((1/( (√2)))−(5/(3(√6))))^2 ]^(1/3) )) = ((√2)/( ((((52)/(54))+(5/(3(√3)))))^(1/(3 )) +((((52)/(54))−(5/(3(√3)))))^(1/(3 )) −(1/3))) =((√2)/( (4/3)−(1/3))) = (√2)

$$\mathrm{a}+\mathrm{b}\:=\:\frac{\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} }{\mathrm{a}^{\mathrm{2}} −\mathrm{ab}+\mathrm{b}^{\mathrm{2}} } \\ $$$$\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{5}}{\mathrm{3}\sqrt{\mathrm{6}}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{5}}{\mathrm{3}\sqrt{\mathrm{6}}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$=\:\frac{\sqrt{\mathrm{2}}}{\left[\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{5}}{\mathrm{3}\sqrt{\mathrm{6}}}\right)^{\mathrm{2}} \right]^{\mathrm{1}/\mathrm{3}} −\sqrt[{\mathrm{3}\:}]{\frac{\mathrm{1}}{\mathrm{27}}}\:+\:\left[\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{5}}{\mathrm{3}\sqrt{\mathrm{6}}}\right)^{\mathrm{2}} \right]^{\mathrm{1}/\mathrm{3}} } \\ $$$$=\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt[{\mathrm{3}\:}]{\frac{\mathrm{52}}{\mathrm{54}}+\frac{\mathrm{5}}{\mathrm{3}\sqrt{\mathrm{3}}}}+\sqrt[{\mathrm{3}\:}]{\frac{\mathrm{52}}{\mathrm{54}}−\frac{\mathrm{5}}{\mathrm{3}\sqrt{\mathrm{3}}}}\:−\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\:\frac{\mathrm{4}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}}\:=\:\sqrt{\mathrm{2}} \\ $$

Answered by som(math1967) last updated on 12/May/21

let a=((1/( (√2)))+(5/(3(√6)))) b=((1/( (√2)))−(5/(3(√6)))) a+b=(2/( (√2)))=(√2) ab=((1/2)−((25)/(54)))=(2/(54))=(1/(27)) a^(1/3) b^(1/3) =(1/3) a+b=(√2) (a^(1/3) +b^(1/3) )^3 −3a^(1/3) b^(1/3) (a^(1/3) +b^(1/3) )=(√2) (a^(1/3) +b^(1/3) )^3 −3×(1/3)×(√2)=(√2) (a^(1/3) +b^(1/3) )^3 =2(√2)=((√2))^3 (a^(1/3) +b^(1/3) )=(√2) ∴ ((1/( (√2)))+(5/(3(√6))))^(1/3) +((1/( (√2)))−(5/(3(√6))))^(1/3) =(√2)

$${let}\:{a}=\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{5}}{\mathrm{3}\sqrt{\mathrm{6}}}\right)\:{b}=\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{5}}{\mathrm{3}\sqrt{\mathrm{6}}}\right) \\ $$$${a}+{b}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}} \\ $$$$\:{ab}=\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{25}}{\mathrm{54}}\right)=\frac{\mathrm{2}}{\mathrm{54}}=\frac{\mathrm{1}}{\mathrm{27}} \\ $$$$\:{a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${a}+{b}=\sqrt{\mathrm{2}} \\ $$$$\left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} −\mathrm{3}{a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} \left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)=\sqrt{\mathrm{2}} \\ $$$$\left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} −\mathrm{3}×\frac{\mathrm{1}}{\mathrm{3}}×\sqrt{\mathrm{2}}=\sqrt{\mathrm{2}} \\ $$$$\left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} =\mathrm{2}\sqrt{\mathrm{2}}=\left(\sqrt{\mathrm{2}}\right)^{\mathrm{3}} \\ $$$$\left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)=\sqrt{\mathrm{2}} \\ $$$$\therefore\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{5}}{\mathrm{3}\sqrt{\mathrm{6}}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{5}}{\mathrm{3}\sqrt{\mathrm{6}}}\right)^{\mathrm{1}/\mathrm{3}} =\sqrt{\mathrm{2}} \\ $$$$ \\ $$

Answered by mnjuly1970 last updated on 12/May/21

X^3 :=(√2) +3((((1/2)−((25)/(54))))^(1/3) )(X) X^3 =(√2) +3(((27−25)/(54)))^(1/3) X X^3 =(√2) +X⇒ X=(√2) ∈R X^3 −2X+X−(√2)=0 X(X−(√2) )(X+(√2) )+(X−(√2) )=0 ⇒ {_(X^2 +X(√2) +1=0 ⇒X∈C) ^( X=(√2))

$${X}^{\mathrm{3}} :=\sqrt{\mathrm{2}}\:+\mathrm{3}\left(\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{25}}{\mathrm{54}}}\:\right)\left({X}\right) \\ $$$$\:\:\:{X}^{\mathrm{3}} =\sqrt{\mathrm{2}}\:+\mathrm{3}\sqrt[{\mathrm{3}}]{\frac{\mathrm{27}−\mathrm{25}}{\mathrm{54}}}\:{X} \\ $$$$\:\:\:{X}^{\mathrm{3}} =\sqrt{\mathrm{2}}\:+{X}\Rightarrow\:{X}=\sqrt{\mathrm{2}}\:\in\mathbb{R}\: \\ $$$$\:\:\:{X}^{\mathrm{3}} −\mathrm{2}{X}+{X}−\sqrt{\mathrm{2}}=\mathrm{0} \\ $$$$\:\:\:\:\:{X}\left({X}−\sqrt{\mathrm{2}}\:\right)\left({X}+\sqrt{\mathrm{2}}\:\right)+\left({X}−\sqrt{\mathrm{2}}\:\right)=\mathrm{0} \\ $$$$\:\:\:\:\Rightarrow\:\left\{_{{X}^{\mathrm{2}} +{X}\sqrt{\mathrm{2}}\:+\mathrm{1}=\mathrm{0}\:\Rightarrow{X}\in\mathbb{C}} ^{\:{X}=\sqrt{\mathrm{2}}} \right. \\ $$$$\:\:\:\:\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com