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Question Number 35256 by bivekverma last updated on 17/May/18

$$\mathrm{Factorize}:{\mathrm{x}}^5-{\mathrm{y}}^5$$

Commented by abdo mathsup 649 cc last updated on 18/May/18

$$letfactorize{x}^n-y^{n}fornintegrletsupposey\ne o$$$$x^n-y^n=y^n({\left(\frac{x}{y}\right)}^n-1)$$$$=y^n(\frac{x}{y}-1)(1+\frac{x}{y}+\frac{x^2}{y^2}+....\frac{x^{n-1}}{y^{n-1}})$$$$=y^{n-1}(x-y)(1+\frac{x}{y}+\frac{x^2}{y^2}+....\frac{x^{n-1}}{y^{n-1}})$$$$=(x-y)(y^{n-1}+{xy}^{n-2}+x^2{y}^{n-3}+...+x^{n-1})for$$$$n=5weget$$$$x^5-y^5=(x-y)(y^4+{xy}^3+x^2{y}^2+x^{3}y+x^4)$$$$=(x-y)(x^4+x^{3}y+x^2{y}^2+{xy}^3+y^4)$$$$weseethat{x}^n-y^n=(x-y)\sum _{i+j=n-1}{x}^i{y}^j$$

Answered by MJS last updated on 17/May/18

$$x^5-y^5=(x-y)(x^4+x^{3}y+x^2{y}^2+{xy}^3+y^4)$$

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