Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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

Last updated date: 20th Jun 2024
Total views: 413.4k
Views today: 10.13k
Verified
413.4k+ views
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.

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 ${\left( {x + y + z} \right)^2}$
$\Rightarrow {\left( {x + y + z} \right)^2} = {\left[ {\left( {2p - 3q} \right) + \left( {p + 2q} \right) + \left( {q - 3p} \right)} \right]^2} \\ \Rightarrow {\left( {x + y + z} \right)^2} = {\left( {2p + p - 3p - 3q + 2q + q} \right)^2} \\$
$\Rightarrow {\left( {x + y + z} \right)^2} = {\left[ {p\left( {2 + 1 - 3} \right) + q\left( { - 3 + 2 + 1} \right)} \right]^2} \\ \Rightarrow {\left( {x + y + z} \right)^2} = {\left[ {p\left( 0 \right) + q\left( 0 \right)} \right]^2} = {0^2} = 0 \\ \therefore {\left( {x + y + z} \right)^2} = 0 \\$
Hence, the value of ${x^2} + {y^2} + {z^2} + 2yz + 2zx + 2xy$ is 0.
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.