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Question Number 41522 by Tawa1 last updated on 09/Aug/18
Find the cube root of 26 − 15(√3)
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{cube}\:\mathrm{root}\:\mathrm{of}\:\:\:\mathrm{26}\:−\:\mathrm{15}\sqrt{\mathrm{3}} \\ $$
Commented by math khazana by abdo last updated on 10/Aug/18
inside C z^3 =26−15(√3) let z=x+iy ⇒ x^3 +3x^2 iy +3x(iy)^2 −iy^3 =26−15(√3) ⇒ x^3 −3xy^2 +i(3x^2 y−y^3 )=26−15(√3) ⇒ x^3 −3xy^2 =26−15(√3) and 3x^2 y −y^3 =0 ⇒3x^2 =y^2 ⇒x^3 −9x^3 =26−15(√3) ⇒−8x^3 =26−15(√3) let find a and b /(a−b(√3))^3 =26−15 (√3) after developpement we get a=2 and b=1 ⇒ (2−(√3))^3 =26−15(√3) ⇒−8x^3 =(2−(√3))^3 ⇒ x^3 =((((√3) −2)/2))^3 ⇒x=((√3)/2) −1 and y =+^− (√3)x =+^− (√3)(((√3)/2)−1) ⇒ z=((√3)/2) −1 +iξ((3/2)−(√3)) with ξ^2 =1
$${inside}\:{C}\:\:{z}^{\mathrm{3}} =\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\:\:{let}\:{z}={x}+{iy}\:\Rightarrow \\ $$$${x}^{\mathrm{3}} \:+\mathrm{3}{x}^{\mathrm{2}} {iy}\:+\mathrm{3}{x}\left({iy}\right)^{\mathrm{2}} \:−{iy}^{\mathrm{3}} \:=\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\:\Rightarrow \\ $$$${x}^{\mathrm{3}} \:−\mathrm{3}{xy}^{\mathrm{2}} \:+{i}\left(\mathrm{3}{x}^{\mathrm{2}} {y}−{y}^{\mathrm{3}} \right)=\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\:\Rightarrow \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{xy}^{\mathrm{2}} \:=\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\:\:{and}\:\mathrm{3}{x}^{\mathrm{2}} {y}\:−{y}^{\mathrm{3}} =\mathrm{0}\:\Rightarrow\mathrm{3}{x}^{\mathrm{2}} ={y}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{3}} \:−\mathrm{9}{x}^{\mathrm{3}} \:=\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\:\Rightarrow−\mathrm{8}{x}^{\mathrm{3}} \:=\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}} \\ $$$${let}\:{find}\:{a}\:{and}\:{b}\:/\left({a}−{b}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} \:=\mathrm{26}−\mathrm{15}\:\sqrt{\mathrm{3}}\:{after} \\ $$$${developpement}\:{we}\:{get}\:\:{a}=\mathrm{2}\:{and}\:{b}=\mathrm{1}\:\Rightarrow \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{3}} \:=\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\:\Rightarrow−\mathrm{8}{x}^{\mathrm{3}} \:=\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{3}} \:\Rightarrow \\ $$$${x}^{\mathrm{3}} \:=\left(\frac{\sqrt{\mathrm{3}}\:−\mathrm{2}}{\mathrm{2}}\right)^{\mathrm{3}} \:\Rightarrow{x}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:−\mathrm{1}\:{and}\:\:{y}\:=\overset{−} {+}\sqrt{\mathrm{3}}{x} \\ $$$$=\overset{−} {+}\sqrt{\mathrm{3}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\mathrm{1}\right)\:\Rightarrow \\ $$$${z}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:−\mathrm{1}\:+{i}\xi\left(\frac{\mathrm{3}}{\mathrm{2}}−\sqrt{\mathrm{3}}\right)\:\:{with}\:\xi^{\mathrm{2}} =\mathrm{1} \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa1 last updated on 10/Aug/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18
a^3 −3a^2 b+3ab^2 −b^3 2^3 −3.2^2 .(√3) +3.2.((√3) )^2 −((√3) )^3 =(2−(√3) )^3 so cube root is (2−(√3) )
$${a}^{\mathrm{3}} −\mathrm{3}{a}^{\mathrm{2}} {b}+\mathrm{3}{ab}^{\mathrm{2}} −{b}^{\mathrm{3}} \\ $$$$\mathrm{2}^{\mathrm{3}} −\mathrm{3}.\mathrm{2}^{\mathrm{2}} .\sqrt{\mathrm{3}}\:+\mathrm{3}.\mathrm{2}.\left(\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{3}}\:\right)^{\mathrm{3}} \\ $$$$=\left(\mathrm{2}−\sqrt{\mathrm{3}}\:\right)^{\mathrm{3}} \\ $$$${so}\:{cube}\:{root}\:{is}\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\:\right) \\ $$
Commented by Joel578 last updated on 09/Aug/18
How did u know a = 2 and b = (√3) ?
$$\mathrm{How}\:\mathrm{did}\:\mathrm{u}\:\mathrm{know}\:{a}\:=\:\mathrm{2}\:\mathrm{and}\:{b}\:=\:\sqrt{\mathrm{3}}\:? \\ $$
Commented by $@ty@m last updated on 09/Aug/18
−3a^2 b−b^3 =−15(√3) it is possible only if b=(√3) ⇒3a^2 (√3)+3(√3)=15(√3) ⇒3a^2 (√3)=12(√3) ⇒a=2
$$−\mathrm{3}{a}^{\mathrm{2}} {b}−{b}^{\mathrm{3}} =−\mathrm{15}\sqrt{\mathrm{3}} \\ $$$${it}\:{is}\:{possible}\:{only}\:{if}\:{b}=\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{3}{a}^{\mathrm{2}} \sqrt{\mathrm{3}}+\mathrm{3}\sqrt{\mathrm{3}}=\mathrm{15}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{3}{a}^{\mathrm{2}} \sqrt{\mathrm{3}}=\mathrm{12}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{a}=\mathrm{2} \\ $$
Commented by Joel578 last updated on 09/Aug/18
understood. thank you very much
$${understood}.\:{thank}\:{you}\:{very}\:{much} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18
or method 26−15(√3) =(a−b(√3) )^3 a^3 −3a^2 b(√3) +3a.b^2 .3−3(√3) b^3 =26−15(√3) (a^3 +9ab^2 )−(√3) (3a^2 b+3b^3 )=26−15(√(3 )) so a^3 +9ab^2 =26 3a^2 b+3b^3 =15 now solve to find a and b b(a^2 +b^2 )=5 a(a^2 +9b^2 )=26 ((a(a^2 +9b^2 )/(b(a^2 +b^2 )))=((26)/5) (a/b)×((((a^2 /b^2 ))+9)/(((a^2 /b^2 ))+1))=((26)/5) let k=(a/b) k.((k^2 +9)/(k^2 +1))=((26)/5) 5(k^3 +9k)=26k^2 +26 5k^3 −26k^2 +45k−26=0 5k^3 −10k^2 −16k^2 +32k+13k−26=0 5k^2 (k−2) −16k (k−2) +13 (k−2)=0 (k−2)(5k^2 −16k+13)=0 so k=2 that means(a/b)=2 b(a^2 +b^2 )=5 b(4b^2 +b^2 )=5 5b^3 =1 so b=1 a=2b that means a=2×1=2 26−15(√3) =(a−b(√3) )^3 26−15(√3) =(2−(√3) )^3
$${or}\:{method} \\ $$$$\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\:=\left({a}−{b}\sqrt{\mathrm{3}}\:\right)^{\mathrm{3}} \\ $$$${a}^{\mathrm{3}} −\mathrm{3}{a}^{\mathrm{2}} {b}\sqrt{\mathrm{3}}\:+\mathrm{3}{a}.{b}^{\mathrm{2}} .\mathrm{3}−\mathrm{3}\sqrt{\mathrm{3}}\:{b}^{\mathrm{3}} =\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\: \\ $$$$\left({a}^{\mathrm{3}} +\mathrm{9}{ab}^{\mathrm{2}} \right)−\sqrt{\mathrm{3}}\:\left(\mathrm{3}{a}^{\mathrm{2}} {b}+\mathrm{3}{b}^{\mathrm{3}} \right)=\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}\:} \\ $$$${so}\:{a}^{\mathrm{3}} +\mathrm{9}{ab}^{\mathrm{2}} =\mathrm{26} \\ $$$$\mathrm{3}{a}^{\mathrm{2}} {b}+\mathrm{3}{b}^{\mathrm{3}} =\mathrm{15} \\ $$$${now}\:{solve}\:{to}\:{find}\:{a}\:{and}\:{b} \\ $$$${b}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)=\mathrm{5} \\ $$$${a}\left({a}^{\mathrm{2}} +\mathrm{9}{b}^{\mathrm{2}} \right)=\mathrm{26} \\ $$$$\frac{{a}\left({a}^{\mathrm{2}} +\mathrm{9}{b}^{\mathrm{2}} \right.}{{b}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}=\frac{\mathrm{26}}{\mathrm{5}} \\ $$$$\frac{{a}}{{b}}×\frac{\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)+\mathrm{9}}{\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)+\mathrm{1}}=\frac{\mathrm{26}}{\mathrm{5}} \\ $$$${let}\:{k}=\frac{{a}}{{b}} \\ $$$${k}.\frac{{k}^{\mathrm{2}} +\mathrm{9}}{{k}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{26}}{\mathrm{5}} \\ $$$$\mathrm{5}\left({k}^{\mathrm{3}} +\mathrm{9}{k}\right)=\mathrm{26}{k}^{\mathrm{2}} +\mathrm{26} \\ $$$$\mathrm{5}{k}^{\mathrm{3}} −\mathrm{26}{k}^{\mathrm{2}} +\mathrm{45}{k}−\mathrm{26}=\mathrm{0} \\ $$$$\mathrm{5}{k}^{\mathrm{3}} −\mathrm{10}{k}^{\mathrm{2}} −\mathrm{16}{k}^{\mathrm{2}} +\mathrm{32}{k}+\mathrm{13}{k}−\mathrm{26}=\mathrm{0} \\ $$$$\mathrm{5}{k}^{\mathrm{2}} \left({k}−\mathrm{2}\right)\:\:−\mathrm{16}{k}\:\:\:\left({k}−\mathrm{2}\right)\:\:+\mathrm{13}\:\:\left({k}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\left({k}−\mathrm{2}\right)\left(\mathrm{5}{k}^{\mathrm{2}} −\mathrm{16}{k}+\mathrm{13}\right)=\mathrm{0} \\ $$$${so}\:{k}=\mathrm{2}\:\:\:{that}\:{means}\frac{{a}}{{b}}=\mathrm{2} \\ $$$${b}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)=\mathrm{5} \\ $$$${b}\left(\mathrm{4}{b}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)=\mathrm{5} \\ $$$$\mathrm{5}{b}^{\mathrm{3}} =\mathrm{1} \\ $$$${so}\:{b}=\mathrm{1} \\ $$$${a}=\mathrm{2}{b}\:\:{that}\:{means}\:{a}=\mathrm{2}×\mathrm{1}=\mathrm{2} \\ $$$$\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\:=\left({a}−{b}\sqrt{\mathrm{3}}\:\right)^{\mathrm{3}} \\ $$$$\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\:=\left(\mathrm{2}−\sqrt{\mathrm{3}}\:\right)^{\mathrm{3}} \\ $$$$ \\ $$
Commented by Tawa1 last updated on 09/Aug/18
Wow, God bless you sir.
$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18
thank u...
$${thank}\:{u}… \\ $$
Commented by Joel578 last updated on 09/Aug/18
Just a little mistake Sir 5b^3 = 5 → b^3 = 1 → b = 1
$$\mathrm{Just}\:\mathrm{a}\:\mathrm{little}\:\mathrm{mistake}\:\mathrm{Sir} \\ $$$$\mathrm{5}{b}^{\mathrm{3}} \:=\:\mathrm{5}\:\:\rightarrow\:{b}^{\mathrm{3}} \:=\:\mathrm{1}\:\:\rightarrow\:{b}\:=\:\mathrm{1} \\ $$

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