The cube of a trinomial is a very useful algebraic identity in mathematics. Instead of multiplying a + b + c three times manually, we use a direct expansion formula that saves time and prevents errors. This shortcut is widely used in algebraic simplification, polynomial identities, and quick calculations in competitive exams. Understanding the formula for (a + b + c)³ helps students solve higher-level maths problems faster in SSC, CUET, JEE, Banking, and school exams.
Formula for Calculating (a + b + c)³ – Overview
| Formula | Variables | When It Is Used |
|---|---|---|
| (a + b + c)³ = a³ + b³ + c³ + 3a²b + 3a²c + 3b²a + 3b²c + 3c²a + 3c²b + 6abc | a, b, c are numbers or algebraic variables | Used in algebraic expansion, polynomial simplification, factorization & identities |
What is (a + b + c)³ in Maths?
In mathematics, (a + b + c)³ represents the cube of a trinomial. To avoid long manual multiplication like
(a + b + c)(a + b + c)(a + b + c),
we use a direct identity that gives the expanded form instantly.
The expression means multiplying the trinomial three times. When expanded, many like terms appear, which are then simplified to form the final identity.
How the expansion works:
Step 1: Expand (a + b + c)(a + b + c)
= a² + b² + c² + 2ab + 2ac + 2bc
Step 2: Multiply the result by the third bracket.
Step 3: Combine all like terms.
This gives the final expanded identity:
a³ + b³ + c³ + 3a²b + 3a²c + 3b²a + 3b²c + 3c²a + 3c²b + 6abc
This formula is extremely important for solving algebraic identities, polynomial simplification, and shortcut maths in exams.
Examples to Calculate (a + b + c)³
Example 1: Expand (x + y + 1)³
Step 1: Use formula → (a + b + c)³
Step 2: Let a = x, b = y, c = 1
Now expand using the identity:
= x³ + y³ + 1³
- 3x²y + 3x²(1)
- 3y²x + 3y²(1)
- 3(1)²x + 3(1)²y
- 6xy(1)
Now simplify all terms:
= x³ + y³ + 1
- 3x²y + 3y²x + 6xy
- 3x² + 3y² + 3x + 3y
Final Answer:
(x + y + 1)³ = x³ + y³ + 1 + 3x²y + 3y²x + 6xy + 3x² + 3y² + 3x + 3y
Example 2: Expand (2 + 1 + 3)³
Step 1: a = 2, b = 1, c = 3
(2 + 1 + 3) = 6
So, (2 + 1 + 3)³ = 6³ = 216
Now expand using identity:
= 2³ + 1³ + 3³
- 3(2²)(1) + 3(2²)(3)
- 3(1²)(2) + 3(1²)(3)
- 3(3²)(2) + 3(3²)(1)
- 6(2)(1)(3)
After simplification, all terms add up to 216.
Final Answer:
(2 + 1 + 3)³ = 216
FAQs about (a + b + c)³ Formula
Q1. Is (a + b + c)³ the same as (a + b)³?
No, (a + b)³ applies to a binomial, while (a + b + c)³ is for a trinomial.
Q2. Can negative values be used in (a + b + c)³?
Yes, the identity works for both positive and negative numbers.
Q3. Where is (a + b + c)³ asked in exams?
It appears in JEE, CUET, SSC, Banking exams, and school-level algebra.
Q4. Is there a shorter compact version of the formula?
Yes: (a + b + c)³ = a³ + b³ + c³ + 3(a + b)(b + c)(c + a).
Q5. Is this formula useful in real life?
Yes, it is used in polynomial modeling, computations, and algebraic simplification.