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Question Number 43707 by Tawa1 last updated on 14/Sep/18

$$\mathrm{Simplify}:$$$$(\mathrm{x}+\mathrm{y}+\mathrm{z})({\mathrm{x}}^{-1}+{\mathrm{y}}^{-1}+{\mathrm{z}}^{-1})=\left({\mathrm{x}}^{-1}{\mathrm{y}}^{-1}{\mathrm{z}}^{-1}\right)(\mathrm{x}+\mathrm{y})(\mathrm{y}+\mathrm{z})(\mathrm{z}+\mathrm{x})$$

Commented by math1967 last updated on 14/Sep/18

$$Simplify???/prove??$$

Commented by Tawa1 last updated on 14/Sep/18

$$\mathrm{It}\mathrm{should}\mathrm{be}\mathrm{prove}\mathrm{sir}$$

Commented by Tawa1 last updated on 14/Sep/18

$$\mathrm{thanks}$$

Answered by math1967 last updated on 14/Sep/18

$$IthinkL.H.S$$$$=(x+y+z)(x^{-1}+y^{-1}+z^{-1})-1$$$$=(x+y+z)\left(\frac{yz+zx+xy}{xyz}\right)-1$$$$=\frac{(x+y+z)(xy+yz+zx)-xyz}{xyz}$$$$=\frac{z{(x+y)}^2+xy(x+y)+z^2(x+y)+xyz-xyz}{xyz}$$$$=\frac{(x+y)(zx+zy+xy+z^2)}{xyz}$$$$=\frac{(x+y)(y+z)(z+x)}{xyz}$$$$=\left(\frac{1}{xyz}\right)(x+y)(y+z)(z+x)$$$$=\left(x^{-1}{y}^{-1}{z}^{-1}\right)(x+y)(y+z)(z+x)$$$$=R.H.S$$

Commented by Tawa1 last updated on 14/Sep/18

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

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