(a + b + c)² Identity
The (a + b + c)² identity extends the familiar (a + b)² identity to three variables. It is used to expand the square of a trinomial (three-term expression).
This identity is essential for simplifying algebraic expressions, verifying numerical calculations, and solving problems in Class 8 and Class 9 mathematics.
The expansion produces six terms — three squares and three cross-product terms (each multiplied by 2).
What is (a + b + c)² Identity?
Identity:
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
This means:
- The square of a trinomial equals the sum of squares of each term plus twice the product of every pair of terms.
- There are C(3,2) = 3 pairs: (a,b), (b,c), (c,a).
- The result has 6 terms in total.
(a + b + c)² Identity Formula
Related formulas:
- (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
- (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
- If a + b + c = s and a² + b² + c² = q, then ab + bc + ca = (s² − q)/2.
- If a + b + c and ab + bc + ca are known, then a² + b² + c² = (a+b+c)² − 2(ab+bc+ca).
Special case with subtraction:
- (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
Derivation and Proof
Proof of (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:
- Write (a + b + c)² = (a + b + c)(a + b + c).
- Let (b + c) = d. Then (a + d)² = a² + 2ad + d².
- Substitute d = b + c:
= a² + 2a(b + c) + (b + c)² - Expand (b + c)² = b² + 2bc + c².
- Expand 2a(b + c) = 2ab + 2ac.
- Combine: = a² + 2ab + 2ac + b² + 2bc + c²
- Rearrange: = a² + b² + c² + 2ab + 2bc + 2ca
The identity is proved.
Types and Properties
Types of problems:
- Type 1 — Direct expansion: Expand (2x + 3y + 4z)².
- Type 2 — Finding a² + b² + c²: Given a + b + c and ab + bc + ca, find a² + b² + c².
- Type 3 — Finding ab + bc + ca: Given a + b + c and a² + b² + c², find ab + bc + ca.
- Type 4 — Numerical verification: Find 31² using (30 + 1)² or (20 + 10 + 1)².
- Type 5 — With negative terms: Expand (a − b + c)² by replacing b with −b.
Solved Examples
Example 1: Example 1: Direct expansion
Problem: Expand (x + 2y + 3z)².
Solution:
Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:
Here a = x, b = 2y, c = 3z.
- a² = x²
- b² = 4y²
- c² = 9z²
- 2ab = 2(x)(2y) = 4xy
- 2bc = 2(2y)(3z) = 12yz
- 2ca = 2(3z)(x) = 6xz
Answer: (x + 2y + 3z)² = x² + 4y² + 9z² + 4xy + 12yz + 6xz.
Example 2: Example 2: Finding sum of squares
Problem: If a + b + c = 10 and ab + bc + ca = 31, find a² + b² + c².
Solution:
Using: (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
- 10² = a² + b² + c² + 2(31)
- 100 = a² + b² + c² + 62
- a² + b² + c² = 100 − 62 = 38
Answer: a² + b² + c² = 38.
Example 3: Example 3: Finding sum of products
Problem: If a + b + c = 9 and a² + b² + c² = 35, find ab + bc + ca.
Solution:
- (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
- 81 = 35 + 2(ab + bc + ca)
- 2(ab + bc + ca) = 46
- ab + bc + ca = 23
Answer: ab + bc + ca = 23.
Example 4: Example 4: Numerical calculation
Problem: Find 111² using the identity.
Solution:
111 = 100 + 10 + 1. So a = 100, b = 10, c = 1.
- (100 + 10 + 1)² = 100² + 10² + 1² + 2(100)(10) + 2(10)(1) + 2(1)(100)
- = 10000 + 100 + 1 + 2000 + 20 + 200
- = 12321
Answer: 111² = 12321.
Example 5: Example 5: With negative term
Problem: Expand (2a − 3b + c)².
Solution:
Replace b by −3b in the identity. Here the three terms are 2a, −3b, c.
- (2a)² + (−3b)² + c² + 2(2a)(−3b) + 2(−3b)(c) + 2(c)(2a)
- = 4a² + 9b² + c² − 12ab − 6bc + 4ca
Answer: (2a − 3b + c)² = 4a² + 9b² + c² − 12ab − 6bc + 4ca.
Example 6: Example 6: Verify numerically
Problem: Verify (1 + 2 + 3)² = 1² + 2² + 3² + 2(1)(2) + 2(2)(3) + 2(3)(1).
Solution:
- LHS = 6² = 36
- RHS = 1 + 4 + 9 + 4 + 12 + 6 = 36
LHS = RHS ✓
Answer: Verified. Both sides equal 36.
Example 7: Example 7: Application to factorisation
Problem: If x² + y² + z² + 2xy + 2yz + 2zx = 49, find x + y + z.
Solution:
The LHS is the expansion of (x + y + z)².
- (x + y + z)² = 49
- x + y + z = ±7
Answer: x + y + z = 7 or −7.
Example 8: Example 8: Three-digit square
Problem: Find 203² using the identity.
Solution:
203 = 200 + 3 (two terms, but we can use the simpler (a+b)² here). Alternatively, 203 = 200 + 0 + 3.
- (200 + 3)² = 200² + 2(200)(3) + 3²
- = 40000 + 1200 + 9
- = 41209
Answer: 203² = 41209.
Example 9: Example 9: Finding individual values
Problem: If a + b + c = 12, a² + b² + c² = 50, and a = 2, find bc.
Solution:
Step 1: b + c = 12 − 2 = 10.
Step 2: b² + c² = 50 − 4 = 46.
Step 3: (b + c)² = b² + c² + 2bc → 100 = 46 + 2bc → 2bc = 54 → bc = 27.
Answer: bc = 27.
Example 10: Example 10: All negative terms
Problem: Expand (−a − b − c)².
Solution:
- (−a − b − c)² = [−(a + b + c)]² = (a + b + c)²
- = a² + b² + c² + 2ab + 2bc + 2ca
Squaring a negative expression gives the same result as squaring the positive version.
Answer: (−a − b − c)² = a² + b² + c² + 2ab + 2bc + 2ca.
Real-World Applications
Applications of this identity:
- Quick calculations: Computing squares of three-digit numbers like 111² = 12321.
- Algebraic simplification: Expanding and simplifying trinomial squares.
- Finding unknown sums: If (a+b+c) and one of a²+b²+c² or ab+bc+ca are known, find the other.
- Geometry: Computing areas and diagonal lengths involving sums of three quantities.
- Physics: Error propagation when three measurements are added.
Key Points to Remember
- (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
- The expansion has 3 square terms and 3 cross-product terms.
- a² + b² + c² = (a + b + c)² − 2(ab + bc + ca).
- ab + bc + ca = [(a + b + c)² − (a² + b² + c²)] / 2.
- For subtraction, change the sign of the relevant term before applying the identity.
- (−a − b − c)² = (a + b + c)² (squaring negates the negative).
- This identity extends naturally: (a₁ + a₂ + ... + aₙ)² = Σaᵢ² + 2Σᵢ<ⱼ aᵢaⱼ.
- Always double-check the signs of the cross-product terms.
Practice Problems
- Expand (3x + y + 2z)².
- If a + b + c = 15 and a² + b² + c² = 83, find ab + bc + ca.
- Find 112² using the identity with 100 + 10 + 2.
- Expand (x − y + 2z)².
- If a + b + c = 0, prove that a² + b² + c² = −2(ab + bc + ca).
- Expand (a + b − c)² and simplify.
- If x + y + z = 8 and xy + yz + zx = 20, find x² + y² + z².
- Factorise: 4x² + y² + 9z² + 4xy + 6yz + 12xz.
Frequently Asked Questions
Q1. What is the (a + b + c)² identity?
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca. It gives the expansion of the square of a sum of three terms.
Q2. How many terms are in the expansion?
Six terms: three squares (a², b², c²) and three cross-products (2ab, 2bc, 2ca).
Q3. How is it derived?
Write (a + b + c) as (a + (b+c)), expand using (x+y)² = x² + 2xy + y², then expand (b+c)² and simplify.
Q4. What happens if one term is negative?
Replace that term with its negative. For (a − b + c)², use a, −b, c in the identity. The cross-products involving −b will be negative.
Q5. How do I find a² + b² + c² if I know a+b+c and ab+bc+ca?
a² + b² + c² = (a + b + c)² − 2(ab + bc + ca). Substitute the known values and calculate.
Q6. What if a + b + c = 0?
Then (a+b+c)² = 0, so a² + b² + c² + 2(ab+bc+ca) = 0, which gives a² + b² + c² = −2(ab+bc+ca).
Q7. Is there a formula for (a + b + c + d)²?
Yes, the same pattern extends: (a+b+c+d)² = a²+b²+c²+d² + 2(ab+ac+ad+bc+bd+cd). Sum of all squares plus twice the sum of all pairwise products.
Q8. Can this identity be used for numerical calculations?
Yes. For example, 321² = (300+20+1)² = 90000 + 400 + 1 + 12000 + 40 + 600 = 103041.
