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Question Number 83941 by john santu last updated on 08/Mar/20

$$\mathrm{If}\sqrt[3]{2}+\sqrt[3]{4}+\sqrt[3]{8}=\mathrm{x}$$$$\mathrm{then}{\mathrm{x}}^3-{6\mathrm{x}}^2+6\mathrm{x}+6=?$$

Answered by $@ty@m123 last updated on 08/Mar/20

$${}^3\sqrt{2}{+}^3\sqrt{4}+2=x$$$${}^3\sqrt{2}{+}^3\sqrt{4}=x-2$$$${\left({}^3\sqrt{2}{+}^3\sqrt{4}\right)}^3={(x-2)}^3$$$$2+4+3.{}^3\sqrt{8}\left({}^3\sqrt{2}{+}^3\sqrt{4}\right)={(x-2)}^3$$$$6+3\times 2\left({}^3\sqrt{2}{+}^3\sqrt{4}\right)={(x-2)}^3$$$$6+6(x-2)={(x-2)}^3$$$$6+6x-12=x^3-8-6x^2+12x$$$$x^3-6x^2+6x+6=8$$

Commented by john santu last updated on 08/Mar/20

$$\mathrm{good}...$$

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