Algebraic Identities and Special Product Formulas
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Other tools you may find usefulShort-cut multiplication formulas - a key tool in algebra
Short-cut multiplication formulas are algebraic identities that allow you to perform calculations on expressions containing variables and numbers more quickly, without having to time-consumingly multiply each term by each ("peer-to-peer"). They are the foundation of mathematics learning at the secondary school level, appearing on high school leaving exams and in everyday engineering and programming calculations. The main uses of these formulas include simplifying complex algebraic expressions, factoring polynomials (factoring), removing irrationalities from the denominator of fractions, and solving quadratic and higher degree equations and inequalities.
Our website presents a complete list of second-degree (squares) and third-degree (cubes) formulas, along with numerical examples and mathematical proofs, which will help you prepare for state tests and exams.
The most important abbreviated second-degree multiplication formulas (Squares)
Second-degree formulas are the most frequently used and involve exponentiating the sums or differences of two expressions:
- Square of the sum:
(a + b)² = a² + 2ab + b²In words:The square of the sum of two terms is equal to the square of the first term plus twice the product of the first and second terms plus the square of the second term.
- Square of the difference:
(a - b)² = a² - 2ab + b²In words:The square of the difference of two terms is equal to the square of the first term minus twice the product of the first and second terms plus the square of the second term.
- Difference of squares:
a² - b² = (a - b)(a + b)In words:The difference of squares of two expressions is equal to the product of the difference of these expressions by their sum.
Third-degree abbreviated multiplication formulas (Cubes)
Third-degree formulas describe geometric and algebraic relationships related to volume (cubes):
- Cube of the sum:
(a + b)³ = a³ + 3a²b + 3ab² + b³ - Cube of difference:
(a - b)³ = a³ - 3a²b + 3ab² - b³ - Sum of cubes:
a³ + b³ = (a + b)(a² - ab + b²) - Difference of cubes:
a³ - b³ = (a - b)(a² + ab + b²)
Table of abbreviated multiplication formulas with examples
The table below collects all formulas in one place and shows their practical application to specific numbers:
| Name of the algebraic formula | Symbolic notation of the formula | Practical calculation example | Solution and final result |
|---|---|---|---|
| Square of the sum | (a + b)² = a² + 2ab + b² | (x + 3)² | x² + 6x + 9 |
| Square of difference | (a - b)² = a² - 2ab + b² | (2x - 5)² | 4x² - 20x + 25 |
| Difference of squares | a² - b² = (a - b)(a + b) | x² - 16 | (x - 4)(x + 4) |
| Cube of the sum | (a + b)³ = a³ + 3a²b + 3ab² + b³ | (x + 2)³ | x³ + 6x² + 12x + 8 |
| Cube of difference | (a - b)³ = a³ - 3a²b + 3ab² - b³ | (x - 1)³ | x³ - 3x² + 3x - 1 |
| Sum of cubes | a³ + b³ = (a + b)(a² - ab + b²) | x³ + 27 | (x + 3)(x² - 3x + 9) |
| Difference cubes | a³ - b³ = (a - b)(a² + ab + b²) | x³ - 8 | (x - 2)(x² + 2x + 4) |
Common mistakes when using the formulas
The most classic mistake made by students is the identification square of the sum with the sum of the squares:
(a + b)² ≠ a² + b²(ERROR! Forgot about the double product 2ab).
To illustrate this in numbers:(3 + 2)² = 5² = 25. Meanwhile, using the wrong formula we would get:3² + 2² = 9 + 4 = 13. The correct formula gives:3² + 2 × 3 × 2 + 2² = 9 + 12 + 4 = 25.
Frequently asked questions (FAQ)
What are abbreviated multiplication formulas?
These are mathematical identities that allow you to transform algebraic expressions more quickly without having to manually multiply each term by each.
What is the formula for the difference of squares?
The formula is: a² - b² = (a - b)(a + b).
Is there a formula for sum of squares in real numbers?
In real numbers, the formula a² + b² does not factor linearly. However, it can be written in complex numbers as (a - bi)(a + bi), where i is an imaginary unit.
How to derive the formula for the square of a sum?
Just multiply two identical brackets by themselves: (a + b)² = (a + b)(a + b) = a² + ab + ba + b² = a² + 2ab + b² (since ab = ba by the commutativity of multiplication).
What are these formulas used for when removing irrationalities from the denominator?
The difference of squares formula is used. By multiplying the numerator and denominator of the fraction by the conjugate expression (e.g. for the denominator √3 - 1 we multiply by √3 + 1), we eliminate the roots in the denominator, obtaining (√3)² - 1² = 3 - 1 = 2.