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Question Number 156419 by cortano last updated on 11/Oct/21

$$\mathrm{x}\mathrm{is}\mathrm{positive}\mathrm{integer}\mathrm{number}$$$$\mathrm{can}\mathrm{you}\mathrm{check}\mathrm{if}\mathrm{Q}=\sqrt[3]{({(\mathrm{x}+2)}^4-{\mathrm{x}}^4}$$$$\mathrm{is}\mathrm{a}\mathrm{natural}\mathrm{number}$$

Commented by mr W last updated on 11/Oct/21

$$x=2:$$$$Q=\sqrt[3]{4^4-2^4}=\sqrt[3]{240}=2\sqrt[3]{30}\notin N$$

Answered by Rasheed.Sindhi last updated on 11/Oct/21

$$\mathrm{Q}=\sqrt[3]{{(\mathrm{x}+2)}^4-{\mathrm{x}}^4}$$$$=\sqrt[3]{{\mathrm{x}}^4+{8\mathrm{x}}^3+{24\mathrm{x}}^2+32\mathrm{x}+16-{\mathrm{x}}^4}$$$$=\sqrt[3]{{8\mathrm{x}}^3+{24\mathrm{x}}^2+32\mathrm{x}+16}$$$$=2\sqrt[3]{{\mathrm{x}}^3+{3\mathrm{x}}^2+4\mathrm{x}+2}$$$$=2\sqrt[3]{({\mathrm{x}}^3+{3\mathrm{x}}^2+3\mathrm{x}+1)+(\mathrm{x}+1)}$$$$=2\sqrt[3]{{(\mathrm{x}+1)}^3+(\mathrm{x}+1)}$$$$\mathrm{The}\mathrm{only}\mathrm{solution},\mathrm{I}\mathrm{think},\mathrm{is}:$$$$\mathrm{x}+1=0$$$$\mathrm{x}=-1\in \mathbb{N}$$$$\mathrm{No}\mathrm{natural}\mathrm{x}$$

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