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Question Number 202447 by maths_plus last updated on 26/Dec/23

$$\mathrm{rationnalise}\mathrm{le}\mathrm{denominateur}\mathrm{de}$$$$\mathrm{x}=\frac{\sqrt[3]{2}-1}{1{-}^3\sqrt{2}{+}^3\sqrt{4}}$$

Answered by cortano12 last updated on 27/Dec/23

$$a^3+b^3=(a+b)(a^2-ab+b^2)$$$$1^3+{\left(\sqrt[3]{2}\right)}^3=(1+\sqrt[3]{2})(1-\sqrt[3]{2}+\sqrt[3]{4})$$$$\frac{1}{3}=\frac{1}{(1+\sqrt[3]{2})(1-\sqrt[3]{2}+\sqrt[3]{4})}$$$$\Rightarrow \mathrm{x}=\sqrt[3]{2}-1.\frac{\sqrt[3]{2}+1}{3}$$

Commented by Frix last updated on 27/Dec/23

$$\mathrm{Yes}\mathrm{but}x\ne \sqrt[3]{2}-1.\frac{\sqrt[3]{2}+1}{3}$$$$x=(\sqrt[3]{2}-1)\frac{\sqrt[3]{2}+1}{3}=-\frac{1-\sqrt[3]{4}}{3}$$

Commented by cortano12 last updated on 27/Dec/23

$$\mathrm{why}?$$

Commented by Frix last updated on 27/Dec/23

$$\mathrm{Sorry}\mathrm{there}\mathrm{was}\mathrm{a}\mathrm{typo}.$$$$\mathrm{But}\mathrm{essentially}\mathrm{you}\mathrm{need}\mathrm{brackets}.$$$$a+b\times \frac{c}{d}\ne (a+b)\times \frac{c}{d}$$

Answered by MATHEMATICSAM last updated on 27/Dec/23

$$1^3+{\left(\sqrt[3]{2}\right)}^3=(\sqrt[3]{2}+1)(1-\sqrt[3]{2}+\sqrt[3]{4})$$$$\mathrm{Now}x=\frac{(\sqrt[3]{2}-1)(\sqrt[3]{2}+1)}{(\sqrt[3]{2}+1)(1-\sqrt[3]{2}+\sqrt[3]{4})}$$$$=\frac{{\left(\sqrt[3]{2}\right)}^2-1}{1+{\left(\sqrt[3]{2}\right)}^3}=\frac{\sqrt[3]{4}-1}{1+2}=\frac{\sqrt[3]{4}-1}{3}\left(\mathrm{Ans}\right)$$

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