Question

If $x + y + z = 9$ and $xy + yz + zx = 9$, then find the value of ${x^2} + {y^2} + {z^2}$.
A. $49$
B. $56$
C. $63$
D. $81$

Answer 1

Hint: In this problem, first we will write the expansion of ${\left( {x + y + z} \right)^2}$. Then, we will substitute the given values in this expansion. Then, we will simplify the expression to get required value.

Complete step-by-step solution:
In this problem, two following equations are given:
$
  x + y + z = 9 \cdots \cdots \left( 1 \right) \\
  xy + yz + zx = 9 \cdots \cdots \left( 2 \right) \\
 $
To find the value of ${x^2} + {y^2} + {z^2}$, first we will write the expansion of ${\left( {x + y + z} \right)^2}$. The expansion of ${\left( {x + y + z} \right)^2}$ is given by ${\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2zx \cdots \cdots \left( 3 \right)$.
Equation $\left( 3 \right)$ can be rewritten as ${\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2\left( {xy + yz + zx} \right) \cdots \cdots \left( 4 \right)$
Here we need to find the value of ${x^2} + {y^2} + {z^2}$. Let us substitute the values from the equation $\left( 1 \right)$ and the equation $\left( 2 \right)$ in equation $\left( 4 \right)$. Therefore, we get
$
  {\left( 9 \right)^2} = {x^2} + {y^2} + {z^2} + 2\left( 9 \right) \\
   \Rightarrow 81 = {x^2} + {y^2} + {z^2} + 18 \\
   \Rightarrow {x^2} + {y^2} + {z^2} = 81 - 18 \\
   \Rightarrow {x^2} + {y^2} + {z^2} = 63 \\
 $
Therefore, if $x + y + z = 9$ and $xy + yz + zx = 9$, then the value of ${x^2} + {y^2} + {z^2}$ is equal to $63$. Therefore, option C is correct.

Note: Let us see how to prove that ${\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2zx$. Expansion means we have to remove parentheses (brackets) in a proper way. To find expansion of ${\left( {x + y + z} \right)^2}$, first we will write ${\left( {x + y + z} \right)^2} = \left( {x + y + z} \right)\left( {x + y + z} \right)$. Now we multiply each term of the first bracket with each term of the second bracket. That is, we will remove parentheses (brackets) in a proper way. Therefore, we get $\left( {x + y + z} \right)\left( {x + y + z} \right) = x\left( {x + y + z} \right) + y\left( {x + y + z} \right) + z\left( {x + y + z} \right)$$ \Rightarrow {\left( {x + y + z} \right)^2} = {x^2} + xy + xz + yx + {y^2} + yz + zx + zy + {z^2}$
Now we will combine the equal terms. Therefore, we get
$
  {\left( {x + y + z} \right)^2} = {x^2} + 2xy + 2yz + 2zx + {y^2} + {z^2} \\
   \Rightarrow {\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2zx \\
 $
In this type of problem, if values of $x + y$ and $xy$ are given then we can find the value of ${x^2} + {y^2}$ by using the expansion of ${\left( {x + y} \right)^2}$. The expansion of ${\left( {x + y} \right)^2}$ is given by ${\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy$. Similarly, if values of $x - y$ and $xy$ are given then we can find the value of ${x^2} + {y^2}$ by using the expansion of ${\left( {x - y} \right)^2}$. The expansion of ${\left( {x - y} \right)^2}$ is given by ${\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy$.