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Question Number 123576 by bramlexs22 last updated on 26/Nov/20

Commented by talminator2856791 last updated on 26/Nov/20

the image is not so clear

$$\:\mathrm{the}\:\mathrm{image}\:\mathrm{is}\:\mathrm{not}\:\mathrm{so}\:\mathrm{clear} \\ $$

Commented by benjo_mathlover last updated on 26/Nov/20

(((2)^(1/3) −1))^(1/3) = (a)^(1/3) + (b)^(1/3) + (c)^(1/3) find the value of a+b+c

$$\sqrt[{\mathrm{3}}]{\sqrt[{\mathrm{3}}]{\mathrm{2}}\:−\mathrm{1}}\:=\:\sqrt[{\mathrm{3}}]{{a}}\:+\:\sqrt[{\mathrm{3}}]{{b}}\:+\:\sqrt[{\mathrm{3}}]{{c}}\: \\ $$$${find}\:{the}\:{value}\:{of}\:{a}+{b}+{c} \\ $$

Answered by Dwaipayan Shikari last updated on 26/Nov/20

(((2)^(1/3) −1))^(1/3) =(p)^(1/3) +(c)^(1/3) (2)^(1/3) −1=p+c +3((pc))^(1/3) ((p)^(1/3) +(c)^(1/3) ) (2)^(1/3) −1=p+c+3((pc))^(1/3) ((((2)^(1/3) −1))^(1/3) ) (2)^(1/3) =3((pc))^(1/3) ((((2)^(1/3) −1))^(1/3) ) 2=27pc((2)^(1/3) −1) p+c=−1⇒p+(2/(((2)^(1/3) −1)27p))=−1 p^2 +p+(2/(((2)^(1/3) −1)27))=0 p=((−1±(√(1−(4/(((2)^(1/3) −1)27)))))/2) p=−(1/2)±(1/2)(√(1−(4/(((2)^(1/3) −1)27)))) c=−(1/2)±(1/2)(√(1−(4/(((2)^(1/3) −1)27)))) p=(a+b+3((ab))^(1/3) ((a)^(1/3) +(b)^(1/3) ))

$$\sqrt[{\mathrm{3}}]{\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}}=\sqrt[{\mathrm{3}}]{{p}}+\sqrt[{\mathrm{3}}]{{c}} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}={p}+{c}\:+\mathrm{3}\sqrt[{\mathrm{3}}]{{pc}}\left(\sqrt[{\mathrm{3}}]{{p}}+\sqrt[{\mathrm{3}}]{{c}}\right) \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}={p}+{c}+\mathrm{3}\sqrt[{\mathrm{3}}]{{pc}}\left(\sqrt[{\mathrm{3}}]{\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}}\right) \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{2}}=\mathrm{3}\sqrt[{\mathrm{3}}]{{pc}}\left(\sqrt[{\mathrm{3}}]{\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}}\right) \\ $$$$\mathrm{2}=\mathrm{27}{pc}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right) \\ $$$${p}+{c}=−\mathrm{1}\Rightarrow{p}+\frac{\mathrm{2}}{\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right)\mathrm{27}{p}}=−\mathrm{1} \\ $$$${p}^{\mathrm{2}} +{p}+\frac{\mathrm{2}}{\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right)\mathrm{27}}=\mathrm{0} \\ $$$${p}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}−\frac{\mathrm{4}}{\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right)\mathrm{27}}}}{\mathrm{2}} \\ $$$${p}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}−\frac{\mathrm{4}}{\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right)\mathrm{27}}}\: \\ $$$${c}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}−\frac{\mathrm{4}}{\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right)\mathrm{27}}} \\ $$$${p}=\left({a}+{b}+\mathrm{3}\sqrt[{\mathrm{3}}]{{ab}}\left(\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}\right)\right) \\ $$$$ \\ $$$$ \\ $$

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