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Question Number 159179 by cortano last updated on 14/Nov/21

$${\mathcal{X}}^3-4{\mathcal{X}}^2-6\mathcal{X}-24=0$$

Answered by mr W last updated on 14/Nov/21

$$let\mathcal{X}=x+\frac{4}{3}$$$$x^3+3\times \frac{4}{3}{x}^2+3\times \frac{16}{9}x+\frac{64}{27}-4\times x^2-4\times 2x\times \frac{4}{3}-4\times \frac{16}{9}-6x-6\times \frac{4}{3}-24=0$$$$x^3-\frac{34x}{3}-\frac{992}{27}=0$$$$\mathrm{\Delta }=\sqrt{-\frac{{34}^3}{9^3}+\frac{{496}^2}{{27}^2}}=\frac{2\sqrt{638}}{3}$$$$\Rightarrow x=\sqrt[3]{\frac{2\sqrt{638}}{3}+\frac{496}{27}}-\sqrt[3]{\frac{2\sqrt{638}}{3}-\frac{496}{27}}$$$$\Rightarrow \mathcal{X}=\frac{1}{3}(4+\sqrt[3]{18\sqrt{638}+496}-\sqrt[3]{18\sqrt{638}-496})\approx 5.7635$$

Commented by cortano last updated on 14/Nov/21

$$wawamazing$$

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