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Question Number 21877 by FilupS last updated on 06/Oct/17

$$x{=}^3\sqrt{7+5\sqrt{2}}{+}^3\sqrt{7-5\sqrt{2}}$$$$$$$$1.\mathrm{According}\mathrm{to}\mathrm{a}\mathrm{video},x=2$$$$$$$$2.\mathrm{According}\mathrm{to}\mathrm{WolframAlpha},$$$$x\approx 0.2071+0.3587i\mathrm{for}‘‘\mathrm{principal}{\mathrm{root}}^{″}$$$$\mathrm{and}x=2\sqrt{2}\mathrm{for}‘‘\mathrm{real}-\mathrm{valued}{\mathrm{root}}^{″}$$$$$$$$3.\mathrm{According}\mathrm{to}\mathrm{google},x\approx 2.8284$$$$$$$$\mathrm{Please}\mathrm{help}\mathrm{and}\mathrm{explain}!:)$$

Commented by Joel577 last updated on 06/Oct/17

$${where}^{\prime }sthevideo?$$

Commented by Joel577 last updated on 06/Oct/17

Commented by FilupS last updated on 06/Oct/17

Answered by mrW1 last updated on 06/Oct/17

$$\mathrm{a}{=}^3\sqrt{7+5\sqrt{2}}$$$$\mathrm{b}{=}^3\sqrt{7-5\sqrt{2}}$$$$\mathrm{x}=\mathrm{a}+\mathrm{b}=\mathrm{real}$$$${\mathrm{a}}^3+{\mathrm{b}}^3=14$$$$\mathrm{ab}{=}^3\sqrt{(7+5\sqrt{2})(7-5\sqrt{2})}{=}^3\sqrt{49-50}=-1$$$${\mathrm{a}}^3+{\mathrm{b}}^3=(\mathrm{a}+\mathrm{b})({\mathrm{a}}^2-\mathrm{ab}+{\mathrm{b}}^2)$$$${\mathrm{a}}^3+{\mathrm{b}}^3=(\mathrm{a}+\mathrm{b})({\mathrm{a}}^2+2\mathrm{ab}+{\mathrm{b}}^2-3\mathrm{ab})$$$${\mathrm{a}}^3+{\mathrm{b}}^3=(\mathrm{a}+\mathrm{b})[{(\mathrm{a}+\mathrm{b})}^2-3\mathrm{ab}]$$$$\Rightarrow 14=\mathrm{x}({\mathrm{x}}^2+3)$$$$\Rightarrow {\mathrm{x}}^3+3\mathrm{x}-14=0$$$${\mathrm{x}}^3-{2\mathrm{x}}^2+{2\mathrm{x}}^2-4\mathrm{x}+7\mathrm{x}-14=0$$$$(\mathrm{x}-2)({\mathrm{x}}^2+2\mathrm{x}+7)=0$$$$\Rightarrow \mathrm{x}-2=0$$$$\Rightarrow \mathrm{x}=2$$

Answered by myintkhaing last updated on 06/Oct/17

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