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# How to prove this? If $a + b + c = 0$, then show that, ${a^2} - bc = {b^2} - ca = {c^2} - ab$

Last updated date: 21st Jul 2024
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Hint:
Here, we will use the given condition to find one variable in terms of the other two variables. Then we will take each expression of the equation and substitute the obtained variable in each expression. We will simplify it further and prove that the given statement is true.

Formula Used:
The square of the sum of the numbers is given by an algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$

Complete step by step solution:
The given equation is $a + b + c = 0$.
Now, by rewriting the equation, we get
$\Rightarrow c = - \left( {a + b} \right)$ ………………………………………………………………….$\left( 1 \right)$
Now, we will prove the given condition ${a^2} - bc = {b^2} - ca = {c^2} - ab$ one by one and we will prove that all the three given conditions leads to the same result.
First, we will find the result of the first expression, ${a^2} - bc$, by substituting equation $\left( 1 \right)$. Therefore, we get
${a^2} - bc = {a^2} - b\left( { - \left( {a + b} \right)} \right)$
We know that the product of two negative integers is a positive integer, so we get
$\Rightarrow {a^2} - bc = {a^2} + b\left( {a + b} \right)$
By multiplying the terms, we get
$\Rightarrow {a^2} - bc = {a^2} + {b^2} + ab$ ………………………………………………….$\left( 2 \right)$
Now, we will find the result of the second expression ${b^2} - ca$ by substituting equation $\left( 1 \right)$.
${b^2} - ca = {b^2} - \left( { - \left( {a + b} \right)} \right)a$
We know that the product of two negative integers is a positive integer, so we get
$\Rightarrow {b^2} - ca = {b^2} + \left( {a + b} \right)a$
Multiplying the terms, we get
$\Rightarrow {b^2} - ca = {b^2} + {a^2} + ab$ ………………………………………………….$\left( 3 \right)$
Now, we will find the result of the third expression ${c^2} - ca$ by substituting equation$\left( 1 \right)$.
${c^2} - ca = {\left( { - \left( {a + b} \right)} \right)^2} - ca$
We know that the product of two negative integers is a positive integer.
Now, by using an algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$, we get
$\Rightarrow {c^2} - ab = {a^2} + {b^2} + 2ab - ab$
Simplifying the expression, we get
$\Rightarrow {c^2} - ab = {a^2} + {b^2} + ab$ ………………………………………………….$\left( 4 \right)$
So, we get ${a^2} - bc = {b^2} - ca = {c^2} - ab = {a^2} + {b^2} + ab$

Therefore, ${a^2} - bc = {b^2} - ca = {c^2} - ab$ is proved true when $a + b + c = 0$.

Note:
We know that the given expression is an algebraic expression. An algebraic expression is defined as an expression with the combination of variables, constants and operators. We can also find the variables $a$ and $b$ in terms of the other variable by using the given condition. So, we get $a = - \left( {b + c} \right)$ and $b = - \left( {a + c} \right)$. By substituting these variables in the given expression we will prove the same results which are equal for all the three expressions.