The value of 52 xyz (xy + yz + xz + $y^{2}$ )(z + x) + 104xy (xy + xz + yz + $y^{2}$ ) is ___

\frac{z (z + x)}{2}

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\frac{z (z - x)}{2}

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\frac{x (z - x)}{2}

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\frac{x (z + x)}{2}

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Solution

The correct option is A $\frac{z (z + x)}{2}$
$52 x y z (x y + y z + x z + y^{2})$
$(z + x) \div 104 x y (x y + x z + y a + y^{2})$
$= \frac{52 x y z (x y + y z + x z + y^{2}) (z + x)}{104 x y (x y + x z + y z + y^{2})}$
$= \frac{z (x y + y z + x z + y^{2}) (z + x)}{2 (x y + y z + x z + y^{2})}$
$= \frac{1}{2} (z) (x + z)$
$∴ R e q u i r e d v a l u e i s \frac{z (z + x)}{z}$

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Similar questions

The value of $52 x y z (x y + y z + x z + y^{2}) (z + x) \times 104 x y (x y + x z + y z + y^{2})$ is:

Simplify the expression: \frac{52 w x y z (x y + y z + x z + y^{2}) (z + x)}{104 x y (x y z + x z^{2} + y z^{2} + y^{2} z)}

x^{2} + x y + x z = 135, y^{2} + y z + x y = 351

z^{2} + x z + y z = 243

x^{2} + y^{2} + z^{2} = ?

Simplify the expression: \frac{52 w x y z (x y + y z + x z + y^{2}) (z + x)}{104 x y (x y z + x z^{2} + y z^{2} + y^{2} z)}

If x+y+z=6, xy+xz+yz=11, xyz=6 thenequals

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