Answer

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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 \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}\].

Complete step-by-step answer:

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.

Complete step-by-step answer:

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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