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# Find the value of $\dfrac{{x + y}}{z}$ if ${x^2} + {y^2} + {z^2} - xy - yz - zx = 0$  Hint: First simplify the given term by modifying the terms in form of some algebraic identities. Use the property that the square of a number can only be a positive number later in order to find the answer.

Given equation is: ${x^2} + {y^2} + {z^2} - xy - yz - zx = 0$
We have to find $\dfrac{{x + y}}{z}$
In order to simplify the equation to bring it in the form of some algebraic identity. Let us multiply both the sides of the equation by 2.
$\Rightarrow 2 \times \left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right) = 2 \times \left( 0 \right) \\ \Rightarrow 2{x^2} + 2{y^2} + 2{z^2} - 2xy - 2yz - 2zx = 0 \\$
Let us simplify the terms In order to form the perfect squares on LHS of $\left( {x - y} \right),\left( {y - z} \right)\& \left( {z - x} \right)$
$\Rightarrow {x^2} + {x^2} + {y^2} + {y^2} + {z^2} + {z^2} - 2xy - 2yz - 2zx = 0 \\ \Rightarrow \left( {{x^2} - 2xy + {y^2}} \right) + \left( {{y^2} - 2yz + {z^2}} \right) + \left( {{z^2} - 2zx + {x^2}} \right) = 0 \\$
As we know the algebraic identity
$\left\{ {{a^2} - 2ab + {b^2} = {{\left( {a - b} \right)}^2}} \right\}$
Using the formula in above equation we have
$\Rightarrow {\left( {x - y} \right)^2} + {\left( {y - z} \right)^2} + {\left( {z - x} \right)^2} = 0$
As we know that square of a real number will always be either positive or zero. And in the above equation the sum of three squares is zero. This is only possible when all the individual squares are zero.
$\therefore \left( {x - y} \right) = 0,\left( {y - z} \right) = 0\& \left( {z - x} \right) = 0 \\ \\$
From the above three results we have:
$\because \left( {x - y} \right) = 0 \\ \Rightarrow x = y.........(1) \\ \because \left( {y - z} \right) = 0 \\ \Rightarrow y = z.........(2) \\ \because \left( {z - x} \right) = 0 \\ \Rightarrow z = x.........(3) \\$
Let us assume the value of x=k. From equation (1), (2) and (3) we have
$\Rightarrow x = y = z = k$
To find $\dfrac{{x + y}}{z}$
Let us put the value in the term
$\Rightarrow \dfrac{{x + y}}{z} = \dfrac{{k + k}}{k}{\text{ }}\left[ {\because x = y = z = k} \right] \\ \Rightarrow \dfrac{{x + y}}{z} = \dfrac{{2k}}{k} = 2 \\$
Hence, the value of the term $\dfrac{{x + y}}{z}$ is 2.

Note: In order to solve such problems where we need to find out the value of the function with different unknown variables. The only way to solve the problem is to find the relation between all the unknown variables with the help of the problem statement. Basic algebraic identities used in the problems must be remembered.
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