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Simplify-x-y-z-x-1-y-1-z-1-x-1-y-1-z-1-x-y-y-z-z-x-




Question Number 43707 by Tawa1 last updated on 14/Sep/18
Simplify:     (x + y + z)(x^(−1)  + y^(−1)  + z^(−1) ) = (x^(−1)  y^(−1)  z^(−1) )(x + y)(y + z)(z + x)
$$\mathrm{Simplify}:\:\:\: \\ $$$$\left(\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\right)\left(\mathrm{x}^{−\mathrm{1}} \:+\:\mathrm{y}^{−\mathrm{1}} \:+\:\mathrm{z}^{−\mathrm{1}} \right)\:=\:\left(\mathrm{x}^{−\mathrm{1}} \:\mathrm{y}^{−\mathrm{1}} \:\mathrm{z}^{−\mathrm{1}} \right)\left(\mathrm{x}\:+\:\mathrm{y}\right)\left(\mathrm{y}\:+\:\mathrm{z}\right)\left(\mathrm{z}\:+\:\mathrm{x}\right) \\ $$
Commented by math1967 last updated on 14/Sep/18
Simplify ???/prove??
$${Simplify}\:???/{prove}?? \\ $$
Commented by Tawa1 last updated on 14/Sep/18
It should be prove sir
$$\mathrm{It}\:\mathrm{should}\:\mathrm{be}\:\mathrm{prove}\:\mathrm{sir} \\ $$
Commented by Tawa1 last updated on 14/Sep/18
thanks
$$\mathrm{thanks} \\ $$
Answered by math1967 last updated on 14/Sep/18
I think L.H.S  =(x+y+z)(x^(−1) +y^(−1) +z^(−1) )−1  =(x+y+z)(((yz+zx+xy)/(xyz)))−1  =(((x+y+z)(xy+yz+zx)−xyz)/(xyz))  =((z(x+y)^2 +xy(x+y)+z^2 (x+y)+xyz−xyz)/(xyz))  =(((x+y)(zx+zy+xy+z^2 ))/(xyz))  =(((x+y)(y+z)(z+x))/(xyz))  =((1/(xyz)))(x+y)(y+z)(z+x)  =(x^(−1) y^(−1) z^(−1) )(x+y)(y+z)(z+x)  =R.H.S
$${I}\:{think}\:{L}.{H}.{S} \\ $$$$=\left({x}+{y}+{z}\right)\left({x}^{−\mathrm{1}} +{y}^{−\mathrm{1}} +{z}^{−\mathrm{1}} \right)−\mathrm{1} \\ $$$$=\left({x}+{y}+{z}\right)\left(\frac{{yz}+{zx}+{xy}}{{xyz}}\right)−\mathrm{1} \\ $$$$=\frac{\left({x}+{y}+{z}\right)\left({xy}+{yz}+{zx}\right)−{xyz}}{{xyz}} \\ $$$$=\frac{{z}\left({x}+{y}\right)^{\mathrm{2}} +{xy}\left({x}+{y}\right)+{z}^{\mathrm{2}} \left({x}+{y}\right)+{xyz}−{xyz}}{{xyz}} \\ $$$$=\frac{\left({x}+{y}\right)\left({zx}+{zy}+{xy}+{z}^{\mathrm{2}} \right)}{{xyz}} \\ $$$$=\frac{\left({x}+{y}\right)\left({y}+{z}\right)\left({z}+{x}\right)}{{xyz}} \\ $$$$=\left(\frac{\mathrm{1}}{{xyz}}\right)\left({x}+{y}\right)\left({y}+{z}\right)\left({z}+{x}\right) \\ $$$$=\left({x}^{−\mathrm{1}} {y}^{−\mathrm{1}} {z}^{−\mathrm{1}} \right)\left({x}+{y}\right)\left({y}+{z}\right)\left({z}+{x}\right) \\ $$$$={R}.{H}.{S} \\ $$
Commented by Tawa1 last updated on 14/Sep/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}\: \\ $$

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