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Question Number 97920 by Lekhraj last updated on 10/Jun/20

Commented by 06122004 last updated on 10/Jun/20

$${\mathrm{x}}^3-4\mathrm{x}=15\Rightarrow {\mathrm{x}}^3-{3\mathrm{x}}^2+{3\mathrm{x}}^2-9\mathrm{x}+5\mathrm{x}-15=0\Rightarrow$$$$\Rightarrow (\mathrm{x}-3)({\mathrm{x}}^2+3\mathrm{x}+5)=0\Rightarrow {\mathrm{x}}^2+3\mathrm{x}=-5\Rightarrow$$$$\Rightarrow \mathrm{x}(\mathrm{x}+1)(\mathrm{x}+2)(\mathrm{x}+3)=$$$$=({\mathrm{x}}^2+3\mathrm{x})({\mathrm{x}}^2+3\mathrm{x}+2)=(-5)\ast (-3)=15$$$$\ast \mathrm{Javob}\ast 15$$$$\mathrm{Abdullaxatov}\mathrm{Jasurbek}$$

Commented by bobhans last updated on 10/Jun/20

$$\mathrm{wrong}.$$$$({\mathrm{x}}^2+3\mathrm{x})({\mathrm{x}}^2+3\mathrm{x}+2)=15$$$$\ast \mathrm{bobhans}\ast$$

Commented by Lekhraj last updated on 11/Jun/20

$$correct$$

Commented by Lekhraj last updated on 11/Jun/20

$$why?$$

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