If x+y=18, y+z=24 and z+x=22, then find the value of $(x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x)$ .

1024

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289

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144

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256

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None of these

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Solution

The correct option is A 1024
$x + y = 18 y + z = 24 z + x = 22$
To find $x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x$
$= (x + y + z)^{2} = {(\frac{(2 x + 2 y + 2 z)}{2})}^{2} = {(\frac{(x + y + y + z + z + x)}{2})}^{2} = {(\frac{(18 + 24 + 22)}{2})}^{2} = {(\frac{64}{2})}^{2} = (32)^{2} = 1024$ .

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