Question 12 of 43advanced⚖️ EvaluateTrue / False2 marks

Is this an identity? (a + b - c)^2 + (a - b + c)^2 + (a - b - c)^2 = 2a22a^{2} + 2b22b^{2} + 2c22c^{2}.

Correct Answer

False

Exercise: Exercise Set 4.3 | Q: 4 | (Page 10)
For More Understanding

Explanation

False. The given equation is not an identity because the Left Hand Side (LHS) does not equal the Right Hand Side (RHS) for all values of the variables.

To verify, we expand each term on the LHS using the identity (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca. By substituting negative signs appropriately, we get:

(a+b−c)2=a2+b2+c2+2ab−2bc−2ca(a + b - c)^2 = a^2 + b^2 + c^2 + 2ab - 2bc - 2ca (a−b+c)2=a2+b2+c2−2ab−2bc+2ca(a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ca (a−b−c)2=a2+b2+c2−2ab+2bc−2ca(a - b - c)^2 = a^2 + b^2 + c^2 - 2ab + 2bc - 2ca

Adding these three expressions, the LHS simplifies to 3a2+3b2+3c2−2ab−2bc−2ca3a^2 + 3b^2 + 3c^2 - 2ab - 2bc - 2ca. Since this is not equal to the RHS, 2a2+2b2+2c22a^2 + 2b^2 + 2c^2, the statement is false.

Solution Steps

  1. Step 1: Expand (a+b−c)2(a + b - c)^2 as a2+b2+c2+2ab−2bc−2caa^2 + b^2 + c^2 + 2ab - 2bc - 2ca using the identity (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca.

  2. Step 2: Expand (a−b+c)2(a - b + c)^2 as a2+b2+c2−2ab−2bc+2caa^2 + b^2 + c^2 - 2ab - 2bc + 2ca.

  3. Step 3: Expand (a−b−c)2(a - b - c)^2 as a2+b2+c2−2ab+2bc−2caa^2 + b^2 + c^2 - 2ab + 2bc - 2ca.

  4. Step 4: Add the three expanded expressions to get LHS = 3a2+3b2+3c2−2ab−2bc−2ca3a^2 + 3b^2 + 3c^2 - 2ab - 2bc - 2ca.

  5. Step 5: Compare LHS with RHS (2a2+2b2+2c22a^2 + 2b^2 + 2c^2). Since they are not equal, it is not an identity.

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