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Question Number 52406 by Tawa1 last updated on 07/Jan/19

Find the cube root of 26 + 5(√3)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{cube}\:\mathrm{root}\:\mathrm{of}\:\:\:\:\:\mathrm{26}\:+\:\mathrm{5}\sqrt{\mathrm{3}} \\ $$

Answered by MJS last updated on 07/Jan/19

no exact solution I think it should be 26+15(√3) (a+b(√3))^3 =26+15(√3) (a^3 +9ab^2 )+3(a^2 b+b^3 )(√3)=26+15(√3) a^3 +9ab^2 =26 ⇒ b=((√(26−a^3 ))/(3(√a))) 3(a^2 b+b^3 )=15 ⇒ ⇒ ((2(4a^3 +13)(√(26−a^3 )))/(27(√a^3 )))=5 ⇒ ⇒ a^9 −((39)/2)a^6 +((8085)/(64))a^3 −((2197)/8)=0 a=(t)^(1/3) t^3 −((39)/2)t^2 −((8085)/(64))t−((2197)/8)=0 t=z+((13)/2) z^3 −((27)/(64))z−((351)/(128))=0 D=(p^3 /(27))+(q^2 /4)>0 ⇒ 1 real & 2 complex solutions trying the factors of −((351)/(128)) we find z=(3/2) ⇒ t=8 ⇒ a=2 ⇒ b=1 ((26+15(√3)))^(1/3) =2+(√3)

$$\mathrm{no}\:\mathrm{exact}\:\mathrm{solution} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}} \\ $$$$\left({a}+{b}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} =\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}} \\ $$$$\left({a}^{\mathrm{3}} +\mathrm{9}{ab}^{\mathrm{2}} \right)+\mathrm{3}\left({a}^{\mathrm{2}} {b}+{b}^{\mathrm{3}} \right)\sqrt{\mathrm{3}}=\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}} \\ $$$${a}^{\mathrm{3}} +\mathrm{9}{ab}^{\mathrm{2}} =\mathrm{26}\:\Rightarrow\:{b}=\frac{\sqrt{\mathrm{26}−{a}^{\mathrm{3}} }}{\mathrm{3}\sqrt{{a}}} \\ $$$$\mathrm{3}\left({a}^{\mathrm{2}} {b}+{b}^{\mathrm{3}} \right)=\mathrm{15}\:\Rightarrow \\ $$$$\Rightarrow\:\frac{\mathrm{2}\left(\mathrm{4}{a}^{\mathrm{3}} +\mathrm{13}\right)\sqrt{\mathrm{26}−{a}^{\mathrm{3}} }}{\mathrm{27}\sqrt{{a}^{\mathrm{3}} }}=\mathrm{5}\:\Rightarrow \\ $$$$\Rightarrow\:{a}^{\mathrm{9}} −\frac{\mathrm{39}}{\mathrm{2}}{a}^{\mathrm{6}} +\frac{\mathrm{8085}}{\mathrm{64}}{a}^{\mathrm{3}} −\frac{\mathrm{2197}}{\mathrm{8}}=\mathrm{0} \\ $$$${a}=\sqrt[{\mathrm{3}}]{{t}} \\ $$$${t}^{\mathrm{3}} −\frac{\mathrm{39}}{\mathrm{2}}{t}^{\mathrm{2}} −\frac{\mathrm{8085}}{\mathrm{64}}{t}−\frac{\mathrm{2197}}{\mathrm{8}}=\mathrm{0} \\ $$$${t}={z}+\frac{\mathrm{13}}{\mathrm{2}} \\ $$$${z}^{\mathrm{3}} −\frac{\mathrm{27}}{\mathrm{64}}{z}−\frac{\mathrm{351}}{\mathrm{128}}=\mathrm{0} \\ $$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}>\mathrm{0}\:\Rightarrow\:\mathrm{1}\:\mathrm{real}\:\&\:\mathrm{2}\:\mathrm{complex}\:\mathrm{solutions} \\ $$$$\mathrm{trying}\:\mathrm{the}\:\mathrm{factors}\:\mathrm{of}\:−\frac{\mathrm{351}}{\mathrm{128}}\:\mathrm{we}\:\mathrm{find} \\ $$$${z}=\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow\:{t}=\mathrm{8}\:\Rightarrow\:{a}=\mathrm{2}\:\Rightarrow\:{b}=\mathrm{1} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}}=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$

Commented by Tawa1 last updated on 07/Jan/19

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jan/19

26+15(√3) =((√3) )^3 +12(√3) +26 =((√3) )^3 +3.((√3) )^2 .2 +3.(√3) .(2)^2 +(2)^3 =((√3) +2)^3 cube root is (2+(√3) )

$$\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}\: \\ $$$$=\left(\sqrt{\mathrm{3}}\:\right)^{\mathrm{3}} +\mathrm{12}\sqrt{\mathrm{3}}\:+\mathrm{26} \\ $$$$=\left(\sqrt{\mathrm{3}}\:\right)^{\mathrm{3}} +\mathrm{3}.\left(\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} .\mathrm{2}\:+\mathrm{3}.\sqrt{\mathrm{3}}\:.\left(\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{2}\right)^{\mathrm{3}} \\ $$$$=\left(\sqrt{\mathrm{3}}\:+\mathrm{2}\right)^{\mathrm{3}} \\ $$$${cube}\:{root}\:{is}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right) \\ $$

Commented by Tawa1 last updated on 07/Jan/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by malwaan last updated on 10/Jan/19

great

$$\mathrm{great} \\ $$

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