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Question Number 188512 by vishal1234 last updated on 02/Mar/23

$$x^3+{\mathit{y}}^3+{\mathit{z}}^3-3\mathit{x}\mathit{y}\mathit{z}=$$$$(\mathit{x}+\mathit{y}+\mathit{z})({\mathit{x}}^2+{\mathit{y}}^2+{\mathit{z}}^2-\mathit{x}\mathit{y}-\mathit{y}\mathit{z}-\mathit{z}\mathit{x})$$$$\mathit{i}\mathit{f}\mathit{x}+\mathit{y}+\mathit{z}=0,then$$$$x^3+{\mathit{y}}^3+{\mathit{z}}^3-3\mathit{x}\mathit{y}\mathit{z}$$$$=0\times ({\mathit{x}}^2+{\mathit{y}}^2+{\mathit{z}}^2-\mathit{x}\mathit{y}-\mathit{y}\mathit{z}-\mathit{z}\mathit{x})$$$$=0$$

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