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# Find ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}$ if ${{x}^{2}}+xy+xz=135,{{y}^{2}}+yz+yx=351$ and ${{z}^{2}}+zx+zy=243$.(a) 225(b) 250 (c) 275(d) 300  Hint: Add all three given equations and use the identity ${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx$ to find the value of $\left( x+y+z \right)$. Now, take out $\left( x+y+z \right)$ common from each of the three equations to get x, y and z. And then find the value of ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}$.

We are given that ${{x}^{2}}+xy+xz=135,{{y}^{2}}+yz+yx=351$ and ${{z}^{2}}+zx+zy=243$. Here we have to find the value ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}$.
First of all, let us consider the equations given in the question
${{x}^{2}}+xy+xz=135....\left( i \right)$
${{y}^{2}}+yz+yx=351.....\left( ii \right)$
${{z}^{2}}+zx+zy=243....\left( iii \right)$
Now, let us add equation (i), (ii) and (iii), we get
$\Rightarrow$ ${{x}^{2}}+xy+xz+{{y}^{2}}+yz+yx+{{z}^{2}}+zx+zy=135+351+243$
By rearranging the above equation, we get,
$\Rightarrow$ ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+xy+yx+xz+zx+zy+yz=729$
By adding the like terms in the above equation, we get,
$\Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx=729$
We know that ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca$
By applying this in the above equation we get,
$\Rightarrow {{\left( x+y+z \right)}^{2}}=729$
By substituting $729={{\left( 27 \right)}^{2}}$in the above equation, we get,
${{\left( x+y+z \right)}^{2}}={{\left( 27 \right)}^{2}}$
We know that when ${{a}^{p}}={{b}^{p}}$ then a = b for all values of ‘a’ and ‘b’ except 1 and -1. By applying this in the above equation, we get,
$\Rightarrow$ $x+y+z=27....\left( iv \right)$
Now, by taking out x common from LHS of equation (i), we get,
$\Rightarrow$ $x\left( x+y+z \right)=135$
By substituting the value of $\left( x+y+z \right)$ from equation (iv), we get
$\Rightarrow$ $x\left( 27 \right)=135$
By dividing 27 on both the sides, we get
$\Rightarrow$ $x=5$
By squaring both the sides, we get,
$\Rightarrow$ ${{x}^{2}}={{5}^{2}}=25....\left( a \right)$
Now, by taking out y common from LHS of equation (ii), we get,
$\Rightarrow$ $y\left( y+x+z \right)=351$
By substituting the value of $\left( x+y+z \right)$ from equation (iv), we get,
$\Rightarrow$ $y\left( 27 \right)=351$
By dividing 27 on both the sides, we get,
$\Rightarrow$ $y=13$
By squaring both the sides, we get,
$\Rightarrow$ ${{y}^{2}}=169....\left( b \right)$
Now, by taking out z from LHS of equation (iii), we get,
$\Rightarrow$ $z\left( z+y+x \right)=243$
By substituting the value of $\left( x+y+z \right)$ from equation (iv), we get,
$\Rightarrow$ $z\left( 27 \right)=243$
By dividing 27 on both the sides, we get,
$\Rightarrow$ $z=9$
By squaring both the sides, we get,
$\Rightarrow$ ${{z}^{2}}=81....\left( c \right)$
By adding equation (a), equation (b) and equation (c), we get
$\Rightarrow$ ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}=25+169+81$
$\Rightarrow$ ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}=275$
Hence, we get the value of $\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)$ is 275.

Note: Students should always remember this approach of using the formula ${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx$ whenever they see the terms like ${{x}^{2}},{{y}^{2}},{{z}^{2}}$ and $xy,yz,zx$ together in the same question. Also, students can cross-check their values of x, y and z by substituting it in equation (i), (ii) and (iii) and checking if LHS = RHS or not.
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