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Question Number 164924 by cortano1 last updated on 23/Jan/22

Commented by bobhans last updated on 23/Jan/22

$$\left(1\right){\mathrm{m}}^3=2020+\sqrt{2019}$$$${\mathrm{n}}^3=2017+\sqrt{2019}$$$${\mathrm{m}}^3-{\mathrm{n}}^3=3;3+{\mathrm{n}}^3-{\mathrm{m}}^3=0$$$$\left(2\right)3^3+{\left({\mathrm{n}}^3\right)}^3+{(-{\mathrm{m}}^3)}^3=3(-{3\mathrm{m}}^3{\mathrm{n}}^3)$$$$27+{\mathrm{n}}^9-{\mathrm{m}}^9=-{9\mathrm{m}}^3{\mathrm{n}}^3$$$$27={\mathrm{m}}^9-{9\mathrm{m}}^3{\mathrm{n}}^3-{\mathrm{n}}^9$$

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