Question Number 95982 by mpym last updated on 29/May/20 | ||
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$$\Sigma{x}\left({y}^{\mathrm{3}} −{z}^{\mathrm{3}} \right)=\_\_\_\_\_. \\ $$ | ||
Answered by mr W last updated on 29/May/20 | ||
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$$={x}\left({y}^{\mathrm{3}} −{z}^{\mathrm{3}} \right)+{y}\left({z}^{\mathrm{3}} −{x}^{\mathrm{3}} \right)+{z}\left({x}^{\mathrm{3}} −{y}^{\mathrm{3}} \right) \\ $$$$=−{xy}\left({x}−{y}\right)\left({x}+{y}\right)−{yz}\left({y}−{z}\right)\left({y}+{z}\right)−{zx}\left({z}−{x}\right)\left({z}+{x}\right) \\ $$ | ||