Question

The value of x +y +z if ${x^2} + {y^2} + {z^2} = 18$ and $xy + yz + zx = 9$ is
$
  (a){\text{ 9}} \\
  (b){\text{ 3}} \\
  (c){\text{ 6}} \\
  (d){\text{ 8}} \\
 $

Answer 1

Hint – In this question use the direct algebraic identity that ${\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2\left( {xy + yz + zx} \right)$, substitution of the given values will help evaluating the value of x +y +z.

Complete Step-by-Step solution:
Given data
${x^2} + {y^2} + {z^2} = 18$..................... (1)
And $xy + yz + zx = 9$....................... (2)
Now as we know that
${\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2\left( {xy + yz + zx} \right)$
Now substitute the values in above equation we have,
$ \Rightarrow {\left( {x + y + z} \right)^2} = 18 + 2\left( 9 \right) = 36$
Now take square root on both sides we have,
$ \Rightarrow x + y + z = \sqrt {36} = 6$
So this is the required answer.

Note – We know that ${(x + y)^2} = {x^2} + {y^2} + 2xy$ is a frequently used algebraic identity is obtained using the concept of binomial expansion under binomial theorem , if the same concept is extended to find square of sum of three numbers that we get the formula for ${\left( {x + y + z} \right)^2}$. In this similar fashion sum of cubes of three numbers can also be evaluated.