Is this an identity?
(a + b − c)2 + (a − b + c)2 + (a − b − c)2 = 2a2 + 2b2 + 2c2.
Step-by-step solution
To find: Whether LHS = RHS for every value of a, b, c
Idea: An identity must be true for all values of the letters. Expand each square with (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca (changing signs where needed), add, and compare with the right side. One set of values where the two sides differ (a counter-example) is enough to prove it is not an identity.
- Expand each square (put −c or −b in the three-term identity):
(a + b − c)2 = a2 + b2 + c2 + 2ab − 2bc − 2ca
(a − b + c)2 = a2 + b2 + c2 − 2ab − 2bc + 2ca
(a − b − c)2 = a2 + b2 + c2 − 2ab + 2bc − 2ca1½ marks - Add them. Squares: 3a2 + 3b2 + 3c2.
ab terms: +2 − 2 − 2 = −2ab; bc terms: −2 − 2 + 2 = −2bc; ca terms: −2 + 2 − 2 = −2ca.
LHS = 3a2 + 3b2 + 3c2 − 2ab − 2bc − 2ca½ mark - Compare: LHS − RHS = a2 + b2 + c2 − 2ab − 2bc − 2ca. This is not zero for all values, so the two sides are not always equal.
- Counter-example: take a = 1, b = 0, c = 0.
LHS = 12 + 12 + 12 = 3; RHS = 2 + 0 + 0 = 2. Since 3 ≠ 2, the statement is not an identity.1 mark - Note: it is still an equation that happens to be true for some values. For a = 1, b = 1, c = 0: LHS = 22 + 02 + 02 = 4 and RHS = 2 + 2 + 0 = 4. Being true for a few values does not make it an identity.
Check: Try another set, a = 1, b = 1, c = 1: LHS = 1 + 1 + 1 = 3, RHS = 6. Different again ✓. A related statement that is an identity: if you also add (a + b + c)2, the four squares add up to 4a2 + 4b2 + 4c2.
Answer to write in the exam
(a + b − c)2 = a2 + b2 + c2 + 2ab − 2bc − 2ca [(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
(a − b + c)2 = a2 + b2 + c2 − 2ab − 2bc + 2ca
(a − b − c)2 = a2 + b2 + c2 − 2ab + 2bc − 2ca
LHS = 3a2 + 3b2 + 3c2 − 2ab − 2bc − 2ca
RHS = 2a2 + 2b2 + 2c2
For a = 1, b = 0, c = 0: LHS = 3, RHS = 2, so LHS ≠ RHS
∴ It is not an identity.
Common mistakes that cost marks
- Testing one lucky set of values (such as a = 1, b = 1, c = 0, where both sides are 4) and concluding it is an identity. Agreement for some values proves nothing; one disagreement disproves it.
- Writing (a − b − c)2 with −2bc. Since (−b)(−c) = +bc, the term is +2bc.
- Adding only the square terms (getting 3a2 + 3b2 + 3c2) and forgetting that the cross terms do not all cancel; −2ab − 2bc − 2ca remains.
How this can come in the exam
For a = 2, b = 1, c = 1, the value of (a + b − c)2 + (a − b + c)2 + (a − b − c)2 − (2a2 + 2b2 + 2c2) is
- 0
- 4
- −4
- 8
Show answer
(C) −4
LHS = 22 + 22 + 02 = 8; RHS = 8 + 2 + 2 = 12; 8 − 12 = −4. (Also from a2 + b2 + c2 − 2ab − 2bc − 2ca = 4 + 1 + 1 − 4 − 2 − 4 = −4.)
Show that (a + b + c)2 + (a + b − c)2 + (a − b + c)2 + (a − b − c)2 = 4(a2 + b2 + c2) is an identity.
Show answer
The last three squares add to 3a2 + 3b2 + 3c2 − 2ab − 2bc − 2ca (1 mark). Adding (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca cancels every cross term and leaves 4a2 + 4b2 + 4c2 for all values, so it is an identity (1 mark).Try one yourself
Is (a + b)2 + (a − b)2 = 2a2 + 2b2 an identity? Give a reason.
Show answer
Yes. (a2 + 2ab + b2) + (a2 − 2ab + b2) = 2a2 + 2b2. The ±2ab cancel for every value of a and b, so both sides are always equal.
More questions like this
- Look at the following figure. Justify the identity a2 = (a + b) (a − b) + b2 for yourself.
- 1. Try to evaluate the following using a suitable identity:
(i) 352 (ii) 652 (iii) 852 (iv) 1052
Do you observe any interesting pattern?
2. Observe the two rows of figures below. They represent an algebraic identity. Try to identify it. - Suppose 7x is split as 2x + 5x; can a similar rectangular arrangement be formed? Consider other possibilities and check.
- Algebra tiles can be used to represent products and find factors.
1. Figure out the product of x + 2 and x + 3 using algebra tiles.
2. Lay out algebra tiles for x2 + 11x + 30 in such a way that you will see its factors. - We have seen that (x + 3)(x + 4) = x2 + 7x + 12.
Also (x + 6)(x + 7) = x2 + 13x + 42.
Generalise the pattern to get an expression for (x + a) (x + b).