Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

$$\mathrm{Just}\:\mathrm{a}\:\mathrm{little}\:\mathrm{mistake}\:\mathrm{Sir} \\ $$$$\mathrm{5}{b}^{\mathrm{3}} \:=\:\mathrm{5}\:\:\rightarrow\:{b}^{\mathrm{3}} \:=\:\mathrm{1}\:\:\rightarrow\:{b}\:=\:\mathrm{1} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com