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: Firstly, we will square the equation $(x + y + z) = 9$ and expand the squared equation using the appropriate identity. After that, we will use the second equation $(xy + yz + zx) = 9$ in the expanded equation after which, only ${x^2} + {y^2} + {z^2}$ would be left, after that we will keep ${x^2} + {y^2} + {z^2}$ at LHS and take all other terms to RHS and solve it to get to the final answer.

Complete step-by-step answer:
According to the question, we are given that
$(x + y + z) = 9$ … (1)
$(xy + yz + zx) = 9$ … (2)
Squaring both sides of (1), we get
\[ \Rightarrow {(x + y + z)^2} = {9^2}\] … (3)
We know that,
$ \Rightarrow {(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2ac + 2bc$ … (4)
Hence, using (4) to expand ${(x + y + z)^2}$, we get
$ \Rightarrow {(x + y + z)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2zx$ … (5)
Using (5) and (3), we get
$ \Rightarrow {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2zx = 81$
We know that the distributive property is
$a \times (b + c + d) = ab + ac + ad$
Using the distributive property for opening the brackets, we get
$ \Rightarrow {x^2} + {y^2} + {z^2} + 2(xy + yz + zx) = 81$ … (6)
Using (2) in (6), we get
$ \Rightarrow {x^2} + {y^2} + {z^2} + 2 \times 9 = 81$
On simplification, we get
$ \Rightarrow {x^2} + {y^2} + {z^2} + 18 = 81$
Taking Constant from LHS to RHS in order to solve further, we get
$ \Rightarrow {x^2} + {y^2} + {z^2} = 81 - 18$
On simplification, we get
$ \Rightarrow {x^2} + {y^2} + {z^2} = 63$
Hence, the final answer is C.

Note: The above question is a direct application of the identity ${(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2ac + 2bc$ .
There are many questions with the same format which have a direct application of some basic identities, some examples of those identities are:
1.${a^3} - {b^3} = (a - b)({a^2} + {b^2} + ab)$
2.${a^3} + {b^3} = (a + b)({a^2} + {b^2} - ab)$
3.${(a + b)^3} = {a^3} + {b^3} + 3ab(a + b)$
4. ${(a - b)^3} = {a^3} - {b^3} - 3ab(a - b)$
We must remember these identities to save us time and extra efforts.