Show that x2-y2-z2-xy-yz-zx-12-x-y2-y-z2-z-x2-

x^2+y^2+z^2 xy yz zx=12[(x y)^2+(y z)^2+(z x)^2] First take R.H.S 1/2[(x y)^2+(y z)^2+(z x)^2] We know that (x y)^2=x^2 2xy+y^2 Therefore, 1/ ...

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Factorisation by Common Factors

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Question

Show that $x^{2} + y^{2} + z^{2} - x y - y z - z x = \frac{1}{2} [(x - y)^{2} + (y - z)^{2} + (z - x)^{2}]$ .

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Show that $x^{2} + x y + y^{2}, z^{2} + z x + x^{2}$ and $y^{2} + y z + z^{2}$ are consecutive terms of an A.P., if $x, y$ and $z$ are in A.P.

(a^{2} + b^{2} + c^{2} - b c - c a - a b) (x^{2} + y^{2} + z^{2}

- y z - z x - x y)

= (X^{2} + Y^{2} + Z^{2} - Y Z - Z X - X Y)

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}}

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