Question Number 59102 by Tawa1 last updated on 04/May/19
$$\mathrm{use}\:\mathrm{remainder}\:\mathrm{theorem}\:\mathrm{to}\:\mathrm{factorize}\:\mathrm{completetly}\:\mathrm{the}\:\mathrm{expression} \\ $$$$\:\:\:\:\:\mathrm{x}^{\mathrm{3}} \left(\mathrm{y}\:−\:\mathrm{z}\right)\:+\:\mathrm{y}^{\mathrm{3}} \left(\mathrm{z}\:−\:\mathrm{x}\right)\:+\:\mathrm{z}^{\mathrm{3}} \left(\mathrm{x}\:−\:\mathrm{y}\right) \\ $$
Answered by tanmay last updated on 04/May/19
![E=0 when x=y so (x−y)is factor (E=given expression) similarly (y−z) and (z−x) x^3 (y−z)+y^3 (z−x)+z^3 (x−y) =(x−y)(y−z)(z−x)(symmetric function degree one) =k(x−y)(y−z)(z−x)(x+y+z) now to find value of k in expression E the coefficient of x^3 y=1 in derived expression the coefficuent of x^3 y=−k {kx×y×−x×x→−kx^3 y} so answer =(−1)×(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)×(x−y)(y−z)(z+x)(x+y+z)](https://www.tinkutara.com/question/Q59112.png)
$${E}=\mathrm{0}\:{when}\:{x}={y}\:{so}\:\left({x}−{y}\right){is}\:{factor}\:\left({E}={given}\:{expression}\right) \\ $$$${similarly}\:\left({y}−{z}\right)\:{and}\:\left({z}−{x}\right) \\ $$$${x}^{\mathrm{3}} \left({y}−{z}\right)+{y}^{\mathrm{3}} \left({z}−{x}\right)+{z}^{\mathrm{3}} \left({x}−{y}\right) \\ $$$$=\left({x}−{y}\right)\left({y}−{z}\right)\left({z}−{x}\right)\left({symmetric}\:{function}\:{degree}\:{one}\right) \\ $$$$={k}\left({x}−{y}\right)\left({y}−{z}\right)\left({z}−{x}\right)\left({x}+{y}+{z}\right) \\ $$$${now}\:{to}\:{find}\:{value}\:{of}\:{k} \\ $$$${in}\:{expression}\:{E}\:\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{3}} {y}=\mathrm{1} \\ $$$${in}\:{derived}\:{expression}\:{the}\:{coefficuent}\:{of}\: \\ $$$${x}^{\mathrm{3}} {y}=−{k}\:\:\left\{{kx}×{y}×−{x}×{x}\rightarrow−{kx}^{\mathrm{3}} {y}\right\} \\ $$$${so}\:{answer} \\ $$$$=\left(−\mathrm{1}\right)×\left({x}−{y}\right)\left({y}−{z}\right)\left({z}−{x}\right)\left({x}+{y}+{z}\right) \\ $$$$ \\ $$$${recheck} \\ $$$${x}^{\mathrm{3}} \left({y}−{z}\right)+{y}^{\mathrm{3}} \left({z}−{x}\right)+{z}^{\mathrm{3}} \left({x}−{y}\right) \\ $$$$={x}^{\mathrm{3}} \left({y}−{z}\right)+{yz}\left({y}^{\mathrm{2}} −{z}^{\mathrm{2}} \right)−{x}\left({y}^{\mathrm{3}} −{z}^{\mathrm{3}} \right) \\ $$$$=\left({y}−{z}\right)\left[{x}^{\mathrm{3}} +{y}^{\mathrm{2}} {z}+{yz}^{\mathrm{2}} −{xy}^{\mathrm{2}} −{xyz}−{xz}^{\mathrm{2}} \right) \\ $$$$=\left({y}−{z}\right)\left[−{x}\left({z}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)+{yz}\left({z}−{x}\right)+{y}^{\mathrm{2}} \left({z}−{x}\right)\right] \\ $$$$=\left({y}−{z}\right)\left({z}−{x}\right)\left[−{xz}−{x}^{\mathrm{2}} +{yz}+{y}^{\mathrm{2}} \right] \\ $$$$=\left({y}−{z}\right)\left({z}−{x}\right)\left[−{z}\left({x}−{y}\right)−\left({x}+{y}\right)\left({x}−{y}\right)\right] \\ $$$$=\left({x}−{y}\right)\left({y}−{z}\right)\left({z}−{x}\right)\left(−{x}−{y}−{z}\right) \\ $$$$=\left(−\mathrm{1}\right)×\left({x}−{y}\right)\left({y}−{z}\right)\left({z}+{x}\right)\left({x}+{y}+{z}\right) \\ $$
Commented by Tawa1 last updated on 21/Jul/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$