Notable products

There are some algebraic expressions that because of their importance and use in mathematics, it is worth memorizing. These are called notable products.

Square of the sum

If $a$ and $b$ are real numbers (they can be unknowns!), it is satisfied that:

$(a + b)^{2} = a^{2} + 2 a b + b^{2}$

Example

$(x + 3)^{2} = x^{2} + 2 \cdot 3 \cdot x + 3^{2} = x^{2} + 6 x + 9$

$(2 x + 1)^{2} = (2 x)^{2} + 2 \cdot (2 x) \cdot 1 + 1^{2} = 4 x^{2} + 4 x + 1$

$(x + \frac{1}{2})^{2} = x^{2} + 2 \cdot \frac{1}{2} \cdot x + (\frac{1}{2})^{2} = x^{2} + x + \frac{1}{4}$

Square of the subtraction

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

Example

$(x - 3)^{2} = x^{2} - 2 \cdot 3 \cdot x + 3^{2} = x^{2} - 6 x + 9$

$(2 x - 1)^{2} = (2 x)^{2} - 2 \cdot (2 x) \cdot 1 + 1^{2} = 4 x^{2} - 4 x + 1$

$(x - \frac{1}{2})^{2} = x^{2} - 2 \cdot \frac{1}{2} \cdot x + (\frac{1}{2})^{2} = x^{2} - x + \frac{1}{4}$

Cube of the sum

In the same way as the square, the cube of the sum is also important: $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$

Example

$(x + 3)^{3} = x^{3} + 3 \cdot x^{2} \cdot 3 + 3 \cdot x \cdot 3^{2} + 3^{3} =$

$= x^{3} + 9 x^{2} + 27 x + 27$

$(2 x + 1)^{3} = (2 x)^{3} + 3 \cdot (2 x)^{2} \cdot 1 + 3 \cdot (2 x) \cdot 1^{2} + 1^{3} =$

$= 8 x^{3} + 12 x^{2} + 3 x + 1$

$(x + \frac{1}{2})^{3} = x^{3} + 3 \cdot \frac{1}{2} \cdot x^{2} + 3 (\frac{1}{2})^{2} \cdot x + (\frac{1}{2})^{3} =$

$= x^{3} + \frac{3}{2} x^{2} + \frac{3}{4} x + \frac{1}{8}$

Cube of the subtraction

$(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$

Example

$(x - 3)^{3} = x^{3} - 3 \cdot x^{2} \cdot 3 + 3 \cdot x \cdot 3^{2} - 3^{3} =$

$= x^{3} - 9 x^{2} + 27 x - 27$

$(2 x - 1)^{3} = (2 x)^{3} - 3 \cdot (2 x)^{2} \cdot 1 + 3 \cdot (2 x) \cdot 1^{2} - 1^{3} =$

$= 8 x^{3} - 12 x^{2} + 3 x - 1$

$(x - \frac{1}{2})^{3} = x^{3} - 3 \cdot \frac{1}{2} \cdot x^{2} + 3 (\frac{1}{2})^{2} \cdot x - (\frac{1}{2})^{3} =$

$= x^{3} - \frac{3}{2} x^{2} + \frac{3}{4} x - \frac{1}{8}$

Difference of two squares

When we subtract two squares, the result is the product of the sum by the difference: $a^{2} - b^{2} = (a + b) \cdot (a - b)$

Example

$x^{2} - 9 = x^{2} - 3^{2} = (x - 3) \cdot (x + 3)$

$4 x^{2} - 1 = 2^{2} x^{2} - 1^{2} = (2 x)^{2} - 1^{2} = (2 x + 1) \cdot (2 x - 1)$

$x^{2} - \frac{1}{4} = x^{2} - \frac{1}{2^{2}} = (x - \frac{1}{2}) \cdot (x + \frac{1}{2})$

Difference of two cubes

In a similar way, there is a formula for the difference of the cubes: $a^{3} - b^{3} = (a - b) \cdot (a^{2} + a b + b^{2})$

Example

$x^{3} - 27 = x^{3} - 3^{3} = (x - 3) \cdot (x^{2} + 3 x + 9)$

$8 x^{3} - 1 = 2^{3} x^{3} - 1^{3} = (2 x)^{3} - 1^{3} = (2 x - 1) \cdot (4 x^{2} + 2 x + 1)$

$x^{3} - \frac{1}{8} = x^{3} - \frac{1}{2^{3}} = (x - \frac{1}{2}) \cdot (x^{2} + \frac{x}{2} + \frac{1}{4})$