79.do the following problem with the process of factarisation of polynomials (x-y)+ (y-x)2 is equal to 1)(y -x)2(1-y+x) 2)(x- y)2(1+y+x) 3)(x-y)2(1+y-x) 4)(y-x)2(1+y+x)

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x + y + z = x y z

x (1 - y^{2}) (1 - z^{2}) + y (1 - z^{2}) (1 - x^{2}) + z (1 - x^{2}) (1 - y^{2})

I f x y + y z + z x = 1,

\frac{x}{1 - x^{2}} + \frac{y}{1 - y^{2}} + \frac{z}{1 - z^{2}} = \frac{4 x y z}{(1 - x^{2}) (1 - y^{2}) (1 - z^{2})} .

x y + y z + z x = 1

\frac{x}{1 - x^{2}} + \frac{y}{1 - y^{2}} + \frac{z}{1 - z^{2}} = \frac{4 x y z}{(1 - x^{2}) (1 - y^{2}) (1 - z^{2})}

x y + y z + z x = 1

\frac{x}{1 + x^{2}} + \frac{y}{1 + y^{2}} + \frac{z}{1 + z^{2}} = \frac{2}{[(1 = x^{2}) (1 + y^{2}) (1 + z^{2})]^{1 / 2}}

[\frac{{(x + y)}^{- 1} . {(x - y)}^{- m}}{{(x + y)}^{m + n} . {(x - y)}^{1 + n}}] {(x^{2} - y^{2})}^{1 + m + n}

x \neq y

y \neq 0

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