Question

If $a + b + c = 0$ and ${a^2} + {b^2} + {c^2} = k\left( {{a^2} - bc} \right)$ , then k =
A.0
B.1
C.2
D.3

Answer 1

Hint: We will use the algebraic identity ${(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2(ab + bc + ca)$ to solve this question. We will put the given value of ${a^2} + {b^2} + {c^2} = k\left( {{a^2} - bc} \right)$in this equation and then, we will solve this equation for the value of k and check which option matches the obtained answer.

Complete step-by-step answer:
We are given that $a + b + c = 0$ and ${a^2} + {b^2} + {c^2} = k\left( {{a^2} - bc} \right)$.
We need to calculate the value of k.
We can write $a + b + c = 0$ as $a = - b - c$
Now, if $a + b + c = 0$, then ${(a + b + c)^2} = 0$
$ \Rightarrow {(a + b + c)^2} = 0$
Using the algebraic identity ${(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2(ab + bc + ca)$in the above equation to expand it, we get
$ \Rightarrow {(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2(ab + bc + ca) = 0$
$ \Rightarrow {a^2} + {b^2} + {c^2} + 2(ab + bc + ca) = 0$
Putting the given value of ${a^2} + {b^2} + {c^2} = k\left( {{a^2} - bc} \right)$ in the above equation, we get
$ \Rightarrow k\left( {{a^2} - bc} \right) + 2\left( {ab + bc + ca} \right) = 0$
$ \Rightarrow k\left( {a\left( a \right) - bc} \right) + 2\left( {ab + bc + ca} \right) = 0$
Substituting the value of $a = - b - c$ in above equation, we get
$ \Rightarrow k\left( {a\left( { - b - c} \right) - bc} \right) + 2\left( {ab + bc + ca} \right) = 0$
Simplifying this equation for the value of k, we get
$
   \Rightarrow 2\left( {ab + bc + ca} \right) = - k\left( {a\left( { - b - c} \right) - bc} \right) \\
   \Rightarrow 2\left( {ab + bc + ca} \right) = - k\left( { - ab - ac - bc} \right) \\
   \Rightarrow 2\left( {ab + bc + ca} \right) = k\left( {ab + ac + bc} \right) \\
   \Rightarrow k = 2 \\
 $
Hence, the value of k is found to be 2.
Therefore, option (C) is correct.

Note: In this question, you may go wrong while simplifying the equation after putting the value of the algebraic identity used here as we have to put the value of a derived from the given initial condition to find the value of k. Here, we have used an algebraic identity which can be defined as an equality which holds true for each and every real value of the variables used in the identity whereas an algebraic expression is an equation formed with variables, integer constants and algebraic operations such as addition, multiplication and subtraction, etc.