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Question Number 150979 by mathdanisur last updated on 17/Aug/21

$$\mathrm{if}\mathrm{x}=\sqrt[3]{4}+\sqrt[3]{2}+1\mathrm{find}\frac{3}{\mathrm{x}}+\frac{3}{{\mathrm{x}}^2}+\frac{1}{{\mathrm{x}}^3}=?$$$$$$

Answered by john_santu last updated on 18/Aug/21

$$\mathrm{If}\mathrm{x}=\sqrt[3]{4}+\sqrt[3]{2}+1\mathrm{then}\frac{3}{\mathrm{x}}+\frac{3}{{\mathrm{x}}^2}+\frac{1}{{\mathrm{x}}^3}=?$$$${\mathrm{a}}^2+\mathrm{ab}+{\mathrm{b}}^2=\frac{{\mathrm{a}}^3-{\mathrm{b}}^3}{\mathrm{a}-\mathrm{b}}$$$$\sqrt[3]{2^2}+\sqrt[3]{2}.1+1^2=\frac{2-1}{\sqrt[3]{2}-1}=\frac{1}{\sqrt[3]{2}-1}$$$$\mathrm{x}=\frac{1}{\sqrt[3]{2}-1};\frac{1}{\mathrm{x}}=\sqrt[3]{2}-1$$$$\iff \frac{1}{\mathrm{x}}+1=\sqrt[3]{2};{(\frac{1}{\mathrm{x}}+1)}^3=2$$$$\iff \frac{1}{{\mathrm{x}}^3}+\frac{3}{{\mathrm{x}}^2}+\frac{3}{\mathrm{x}}+1=2$$$$\iff \frac{1}{{\mathrm{x}}^3}+\frac{3}{{\mathrm{x}}^2}+\frac{3}{\mathrm{x}}=1$$

Commented by bramlexs22 last updated on 17/Aug/21

$$\frac{3}{\mathrm{x}}=3\sqrt[3]{2}-3;$$$$\frac{3}{{\mathrm{x}}^2}=3{(\sqrt[3]{2}-1)}^2=3(\sqrt[3]{4}-2\sqrt[3]{2}+1)=3\sqrt[3]{4}-6\sqrt[3]{2}+3$$$$\frac{1}{{\mathrm{x}}^3}={(\sqrt[3]{2}-1)}^3=1-3\sqrt[3]{4}+3\sqrt[3]{2}$$$$\therefore \frac{3}{\mathrm{x}}+\frac{3}{{\mathrm{x}}^2}+\frac{1}{{\mathrm{x}}^3}=\cancel{3\sqrt[3]{2}}-3+\cancel{3\sqrt[3]{4}}-\cancel{6\sqrt[3]{2}}+3+1-\cancel{3\sqrt[3]{4}}+\cancel{3\sqrt[3]{2}}$$$$=-3+3+1=1$$

Commented by mathdanisur last updated on 17/Aug/21

$$\mathrm{Thank}\mathrm{you}\mathrm{Ser}$$

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