Exploring Algebraic Identities | Exercise 4.3

Question 4

Is this an identity?

$(a + b - c)^2 + (a - b + c)^2 + (a - b - c)^2 = 2a^2 + 2b^2 + 2c^2$

Solution

Understand the Question

An algebraic identity is an equation that is true for all values of the variables.
To check if the given statement is an identity, we use the standard algebraic expansion: $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$
We expand all three terms on the Left Hand Side ( $\text{LHS}$ ), combine like terms, and check if the result equals the Right Hand Side ( $\text{RHS} = 2a^2 + 2b^2 + 2c^2$ ).

Step 1 · Expand $(a + b - c)^2$

Using the identity $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$ with $x = a$ , $y = b$ , and $z = -c$ : Diagram 1

\begin{aligned} (a + b - c)^2 &= a^2 + b^2 + (-c)^2 + 2(a)(b) + 2(b)(-c) + 2(a)(-c) \\[0.6em] &= a^2 + b^2 + c^2 + 2ab - 2bc - 2ac \end{aligned}

Step 2 · Expand $(a - b + c)^2$

Applying the identity with $x = a$ , $y = -b$ , and $z = c$ :

\begin{aligned} (a - b + c)^2 &= a^2 + (-b)^2 + c^2 + 2(a)(-b) + 2(-b)(c) + 2(a)(c) \\[0.6em] &= a^2 + b^2 + c^2 - 2ab - 2bc + 2ac \end{aligned}

Step 3 · Expand $(a - b - c)^2$

Applying the identity with $x = a$ , $y = -b$ , and $z = -c$ :

\begin{aligned} (a - b - c)^2 &= a^2 + (-b)^2 + (-c)^2 + 2(a)(-b) + 2(-b)(-c) + 2(a)(-c) \\[0.6em] &= a^2 + b^2 + c^2 - 2ab + 2bc - 2ac \end{aligned}

Step 4 · Add Expanded Terms to Find LHS

Adding the three expanded expressions:

\begin{aligned} \text{LHS} &= (a^2 + b^2 + c^2 + 2ab - 2bc - 2ac) \\ &\quad + (a^2 + b^2 + c^2 - 2ab - 2bc + 2ac) \\ &\quad + (a^2 + b^2 + c^2 - 2ab + 2bc - 2ac) \\[0.6em] &= (a^2 + a^2 + a^2) + (b^2 + b^2 + b^2) + (c^2 + c^2 + c^2) \\ &\quad + (2ab - 2ab - 2ab) + (-2bc - 2bc + 2bc) + (-2ac + 2ac - 2ac) \\[0.6em] &= 3a^2 + 3b^2 + 3c^2 - 2ab - 2bc - 2ac \end{aligned}

Step 5 · Compare LHS with RHS

Given: $\text{RHS} = 2a^2 + 2b^2 + 2c^2$

Comparing $\text{LHS}$ and $\text{RHS}$ : $\text{LHS} = 3a^2 + 3b^2 + 3c^2 - 2ab - 2bc - 2ac \neq \text{RHS}$

For example, setting $a = 1$ , $b = 0$ , $c = 0$ : $\text{LHS} = 3(1)^2 + 3(0)^2 + 3(0)^2 - 2(1)(0) - 2(0)(0) - 2(1)(0) = 3$ $\text{RHS} = 2(1)^2 + 2(0)^2 + 2(0)^2 = 2$

Since $\text{LHS} \neq \text{RHS}$ , the equation does not hold for all values.

Answer

No, it is not an identity.

Common Mistakes

Sign Errors in Expansion: Forgetting that multiplying two negative terms results in a positive term, e.g., $2(-b)(-c) = +2bc$ , not $-2bc$ .
Definition of Identity: An identity must hold for all real values of $a$ , $b$ , and $c$ . If the algebraic expressions of LHS and RHS are not identically equal, it is not an identity.

Question 4

More questions in Exercise 4.3