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x-y-3-z-3-




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

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