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Question Number 165831 by HongKing last updated on 09/Feb/22

$$\mathrm{Find}\mathrm{the}\mathrm{integer}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{number}:$$$$\sqrt[3]{2015\cdot 2016\cdot 2017}$$

Commented by mr W last updated on 09/Feb/22

$$\sqrt[3]{x(x+1)(x+2)}$$$$>\sqrt[3]{x\times x\times x}=x$$$$$$$$\sqrt[3]{x(x+1)(x+2)}=\sqrt[3]{(x+1)({(x+1)}^2-1)}$$$$<\sqrt[3]{(x+1){(x+1)}^2}=x+1$$$$$$$$x<\sqrt[3]{x(x+1)(x+2)}<x+1$$$$\Rightarrow \left[\sqrt[3]{x(x+1)(x+2)}\right]=x$$$$$$$$\Rightarrow [\sqrt[3]{2015\times 2016\times 2017}=2015$$

Answered by naka3546 last updated on 09/Feb/22

$$\mathrm{Let}x=2016$$$$\sqrt[3]{2015\cdot 2016\cdot 2017}$$$$=\sqrt[3]{(x-1)\cdot x\cdot (x+1)}$$$$=\sqrt[3]{x(x^2-1)}$$$$=\sqrt[3]{x^3-x}<\sqrt[3]{x^3}$$$$$$$$Rememberthat$$$$x(x+1)-3x+1<x(x+1)$$$${(x-1)}^2<x(x+1)$$$$$$$$\sqrt[3]{(x-1)\cdot {(x-1)}^2}<\sqrt[3]{(x-1)\cdot x\cdot (x+1)}$$$$2015=\sqrt[3]{{(x-1)}^3}<\sqrt[3]{x\cdot (x^2-1)}<\sqrt[3]{x^3}=2016$$$$$$$$\mathrm{Integer}\mathrm{part}\mathrm{of}\sqrt[3]{2015\cdot 2016\cdot 2017}\mathrm{is}2015.$$

Commented by naka3546 last updated on 09/Feb/22

$$yes,sir.$$

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