Question

If \[x = 2p - 3q,y = p + 2q\] and \[z = q - 3p\], then \[{x^2} + {y^2} + {z^2} + 2yz + 2zx + 2xy\] is
A. 0
B. 1
C. \[xyz\]
D. \[pqr\]

Answer 1

Hint: In this question, we will proceed by writing the given data and using the algebraic identity simplify the polynomial to get the required answer easily. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Given that \[x = 2p - 3q,y = p + 2q\] and \[z = q - 3p\].
We have to find the value of \[{x^2} + {y^2} + {z^2} + 2yz + 2zx + 2xy\]
We know that \[{\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2yz + 2zx + 2xy\]
Now consider the value of \[{\left( {x + y + z} \right)^2}\]
\[
   \Rightarrow {\left( {x + y + z} \right)^2} = {\left[ {\left( {2p - 3q} \right) + \left( {p + 2q} \right) + \left( {q - 3p} \right)} \right]^2} \\
   \Rightarrow {\left( {x + y + z} \right)^2} = {\left( {2p + p - 3p - 3q + 2q + q} \right)^2} \\
\]
Grouping the common terms and simplifying further, we get
\[
   \Rightarrow {\left( {x + y + z} \right)^2} = {\left[ {p\left( {2 + 1 - 3} \right) + q\left( { - 3 + 2 + 1} \right)} \right]^2} \\
   \Rightarrow {\left( {x + y + z} \right)^2} = {\left[ {p\left( 0 \right) + q\left( 0 \right)} \right]^2} = {0^2} = 0 \\
  \therefore {\left( {x + y + z} \right)^2} = 0 \\
\]
Hence, the value of \[{x^2} + {y^2} + {z^2} + 2yz + 2zx + 2xy\] is 0.
Thus, the correct option is A. 0

Note: Here we have used the algebraic identity \[{\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2yz + 2zx + 2xy\] to solve easily. We can directly substitute the given data in \[{x^2} + {y^2} + {z^2} + 2yz + 2zx + 2xy\] in order to get the required answer but it consumes much time. So, use algebraic identities to simplify and solve easily in these kinds of questions.