If $x^{2} + x y + x z = 135, y^{2} + y z + x y = 351$ and $z^{2} + z x + z y = 243$ . Then the value of x is, $(x, y, z > 0)$

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Solution

Given :
$x^{2} + x y + x z = 135$ ..... $(1)$
$y^{2} + y z + x y = 351$ ..... $(2)$
$z^{2} + z x + z y = 243$ ..... $(3)$

Adding eqns $(1)$ , $(2)$ and $(3)$ we get
$x^{2} + x y + x z + y^{2} + y z + x y + z^{2} + z x + z y = 135 + 351 + 243$
$x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x = 729$
${(x + y + z)}^{2} = 729$
$\Rightarrow x + y + z = \sqrt{729} = 27$

From $(1)$
$x^{2} + x y + x z = 135$
$\Rightarrow x (x + y + z) = 135$
$\Rightarrow x \times 27 = 135$
$∴ x = \frac{135}{27} = 5$

Hence, this is the answer.

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