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Question Number 59102 by Tawa1 last updated on 04/May/19

$$\mathrm{use}\mathrm{remainder}\mathrm{theorem}\mathrm{to}\mathrm{factorize}\mathrm{completetly}\mathrm{the}\mathrm{expression}$$$${\mathrm{x}}^3(\mathrm{y}-\mathrm{z})+{\mathrm{y}}^3(\mathrm{z}-\mathrm{x})+{\mathrm{z}}^3(\mathrm{x}-\mathrm{y})$$

Answered by tanmay last updated on 04/May/19

$$E=0whenx=yso(x-y)isfactor(E=givenexpression)$$$$similarly(y-z)and(z-x)$$$$x^3(y-z)+y^3(z-x)+z^3(x-y)$$$$=(x-y)(y-z)(z-x)\left(symmetricfunctiondegreeone\right)$$$$=k(x-y)(y-z)(z-x)(x+y+z)$$$$nowtofindvalueofk$$$$inexpressionEthecoefficientof{x}^{3}y=1$$$$inderivedexpressionthecoefficuentof$$$$x^{3}y=-k\{kx\times y\times -x\times x\to -{kx}^{3}y\}$$$$soanswer$$$$=(-1)\times (x-y)(y-z)(z-x)(x+y+z)$$$$$$$$recheck$$$$x^3(y-z)+y^3(z-x)+z^3(x-y)$$$$=x^3(y-z)+yz(y^2-z^2)-x(y^3-z^3)$$$$=(y-z)[x^3+y^{2}z+{yz}^2-{xy}^2-xyz-{xz}^2)$$$$=(y-z)[-x(z^2-x^2)+yz(z-x)+y^2(z-x)]$$$$=(y-z)(z-x)[-xz-x^2+yz+y^2]$$$$=(y-z)(z-x)[-z(x-y)-(x+y)(x-y)]$$$$=(x-y)(y-z)(z-x)(-x-y-z)$$$$=(-1)\times (x-y)(y-z)(z+x)(x+y+z)$$

Commented by Tawa1 last updated on 21/Jul/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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