Question Number 4963 by ankit chakravarti last updated on 27/Mar/16
$$\Sigma{x}\left({y}^{\mathrm{3}} −{z}^{\mathrm{3}} \right)=…. \\ $$$$ \\ $$
Commented by prakash jain last updated on 27/Mar/16
$${x}\left({y}^{\mathrm{3}} −{z}^{\mathrm{3}} \right)+{y}\left({z}^{\mathrm{3}} −{x}^{\mathrm{3}} \right)+{z}\left({x}^{\mathrm{3}} −{y}^{\mathrm{3}} \right) \\ $$
Commented by prakash jain last updated on 27/Mar/16
$$\mathrm{The}\:\mathrm{highest}\:\mathrm{power}\:\mathrm{is}\:\mathrm{4}\:\mathrm{so}\:\mathrm{it}\:\mathrm{should}\:\mathrm{have} \\ $$$$\mathrm{4}\:\mathrm{factors}. \\ $$$${f}\left({x},{y},{z}\right)={x}\left({y}^{\mathrm{3}} −{z}^{\mathrm{3}} \right)+{y}\left({z}^{\mathrm{3}} −{x}^{\mathrm{3}} \right)+{z}\left({x}^{\mathrm{3}} −{y}^{\mathrm{3}} \right) \\ $$$$\mathrm{If}\:\mathrm{we}\:\mathrm{try}\:{x}={y}\:{f}\left({x},{y},{z}\right)=\mathrm{0}\Rightarrow\left({x}−{y}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{factor}. \\ $$$$\mathrm{Remainder}\:\mathrm{theorem}. \\ $$$$\mathrm{similarly}\:\left({y}−{z}\right)\:\mathrm{and}\:\left({z}−{x}\right)\:\mathrm{are}\:\mathrm{factors}. \\ $$$$\mathrm{Also}\:\left({x}+{y}+{z}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{factor}. \\ $$$${f}\left({x},{y},{z}\right)\:\mathrm{is}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{4}\:\mathrm{so}\:\mathrm{it}\:\mathrm{can}\:\mathrm{have}\:\mathrm{only} \\ $$$$\mathrm{4}\:\mathrm{linear}\:\mathrm{factors}. \\ $$$${f}\left({x},{y},{z}\right)= \\ $$$${x}\left({y}^{\mathrm{3}} −{z}^{\mathrm{3}} \right)+{y}\left({z}^{\mathrm{3}} −{x}^{\mathrm{3}} \right)+{z}\left({x}^{\mathrm{3}} −{y}^{\mathrm{3}} \right)= \\ $$$$\:\:\:\:\left({x}−{y}\right)\left({y}−{z}\right)\left({z}−{x}\right)\left({x}+{y}+{z}\right) \\ $$