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Question Number 90139 by jagoll last updated on 21/Apr/20

$$\mathrm{x}={2021}^3-{2019}^3$$$$\sqrt{\frac{\mathrm{x}-2}{6}}=?$$

Commented by jagoll last updated on 21/Apr/20

$$\mathrm{x}={(\mathrm{p}+2)}^3-{\mathrm{p}}^3,\mathrm{p}=2019$$$$\mathrm{x}={6\mathrm{p}}^2+12\mathrm{p}+8$$$$\mathrm{x}-2={6\mathrm{p}}^2+12\mathrm{p}+6$$$$\mathrm{x}-2=6{(\mathrm{p}+1)}^2$$$$\sqrt{\frac{\mathrm{x}-2}{6}}=\sqrt{\frac{6{(\mathrm{p}+1)}^2}{6}}$$$$=\mathrm{p}+1=2019+1=2020$$

Commented by MJS last updated on 21/Apr/20

$$\mathrm{good}\mathrm{idea},\mathrm{just}\mathrm{modify}\mathrm{it}\mathrm{a}\mathrm{bit}:$$$$t=2020$$$$x={(t+1)}^3-{(t-1)}^3=6t^2+2$$$$x-2=6t^2$$$$\sqrt{\frac{6t^2}{6}}=t=2020$$

Commented by jagoll last updated on 21/Apr/20

$$\mathrm{waw}....\mathrm{very}\mathrm{great}\mathrm{sir}.\mathrm{thank}\mathrm{you}$$

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