Question Number 35256 by bivekverma last updated on 17/May/18
$$\mathrm{Factorize}\::\mathrm{x}^{\mathrm{5}} −\mathrm{y}^{\mathrm{5}} \\ $$
Commented by abdo mathsup 649 cc last updated on 18/May/18
$${let}\:{factorize}\:{x}^{{n}} \:−{y}^{{n}} \:\:{for}\:{n}\:{integr}\:\:{letsuppose}\:{y}\neq{o} \\ $$$${x}^{{n}} \:−{y}^{{n}} ={y}^{{n}} \:\left(\:\:\left(\frac{{x}}{{y}}\right)^{{n}} \:−\mathrm{1}\right) \\ $$$$={y}^{{n}} \left(\:\:\frac{{x}}{{y}}−\mathrm{1}\right)\left(\:\mathrm{1}\:\:+\frac{{x}}{{y}}\:+\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }\:+….\frac{{x}^{{n}−\mathrm{1}} }{{y}^{{n}−\mathrm{1}} }\right) \\ $$$$={y}^{{n}−\mathrm{1}} \left({x}−{y}\right)\left(\:\mathrm{1}+\:\frac{{x}}{{y}}\:+\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }\:+\:….\frac{{x}^{{n}−\mathrm{1}} }{{y}^{{n}−\mathrm{1}} }\right) \\ $$$$=\left({x}−{y}\right)\left(\:{y}^{{n}−\mathrm{1}} \:\:+{xy}^{{n}−\mathrm{2}} \:+{x}^{\mathrm{2}} {y}^{{n}−\mathrm{3}} \:+…+{x}^{{n}−\mathrm{1}} \right)\:{for} \\ $$$${n}=\mathrm{5}\:{we}\:{get} \\ $$$${x}^{\mathrm{5}} \:−{y}^{\mathrm{5}} =\left({x}−{y}\right)\left({y}^{\mathrm{4}} \:+{xy}^{\mathrm{3}} \:+{x}^{\mathrm{2}} \:{y}^{\mathrm{2}} \:+{x}^{\mathrm{3}} {y}\:+{x}^{\mathrm{4}} \right) \\ $$$$=\left({x}−{y}\right)\left({x}^{\mathrm{4}} \:+{x}^{\mathrm{3}} {y}\:+{x}^{\mathrm{2}} {y}^{\mathrm{2}} \:\:+{xy}^{\mathrm{3}} \:+{y}^{\mathrm{4}} \right) \\ $$$${we}\:{see}\:{that}\:{x}^{{n}} −{y}^{{n}} =\left({x}−{y}\right)\sum_{{i}+{j}={n}−\mathrm{1}} {x}^{{i}} {y}^{{j}} \\ $$
Answered by MJS last updated on 17/May/18
$${x}^{\mathrm{5}} −{y}^{\mathrm{5}} =\left({x}−{y}\right)\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} {y}+{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{xy}^{\mathrm{3}} +{y}^{\mathrm{4}} \right) \\ $$