1.12 Special Product Formulas

The following special formulas are vastly used in algebra and calculus, and should be memorized. You can verify each of the formulas by actual multiplication.

1. $(A + B)^{2} = A^{2} + 2 A B + B^{2}$	(Square of a Sum)
2. $(A - B)^{2} = A^{2} - 2 A B + B^{2}$	(Square of a Difference)
3. $(A + B)^{3} = A^{3} + 3 A^{2} B + 3 A B^{2} + B^{3}$	(Cube of a Sum)
4. $(A - B)^{3} = A^{3} - 3 A^{2} B + 3 A B^{2} - B^{3}$	(Cube of a Difference)

Here $A$ and $B$ represent real numbers, variables, or algebraic expressions.

Note than the Square of Difference formula can be obtained if we replace $B$ with $- B$ in the Square of Sum formula. Similarly replacing $B$ with $- B$ in the Cube of Sum formula yields the Cube of Difference formula.

Read more: Expansion of (A ± B)ⁿ

In the above formulas, we reviewed the square and cube of the binomial $A + B$ . The expansion of $(A + B)^{n}$ for any positive integer $n$ is

$(A + B)^{n} = A^{n} + (\binom{n}{1}) A^{n - 1} B + (\binom{n}{2}) A^{n - 2} B + \dots + (\binom{n}{n - 1}) A B + B^{n}$

and

$\begin{aligned} (A - B)^{n} = A^{n} - (\binom{n}{1}) A^{n - 1} B + & \dots + (- 1)^{k} (\binom{n}{k}) A^{n - k} B^{k} \\ + \dots + (- 1)^{n - 1} (\binom{n}{n - 1}) A B^{n - 1} + (- 1)^{n} B^{n} \end{aligned}$

where

$(\binom{n}{k}) = \frac{n!}{k! (n - k)!} (read “n choose k”)$

with $k! = 1 \times 2 \times 3 \times \dots \times (k - 1) \times k .$

For example,

$\begin{aligned} (x + y)^{4} & = x^{4} + (\binom{4}{1}) x^{3} y + (\binom{4}{2}) x^{2} y^{2} + (\binom{4}{3}) x^{3} y + y^{4} \\ = x^{4} + \frac{4!}{1! \times 3!} x^{3} y + \frac{4!}{2! \times 2!} x^{2} y^{2} + \frac{4!}{3! \times 1!} x^{3} y + y^{4} \\ = x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + y^{4} \end{aligned}$

5. $(x + A) (x + B) = x^{2} + (A + B) x + A B$	(Product of two Binomials having a Common Term)
6. $(A - B) (A + B) = A^{2} - B^{2}$	(Product of Sum and Difference)
7. $(A - B) (A^{2} + A B + B^{2}) = A^{3} - B^{3}$	(Difference of Cubes)
8. $(A + B) (A^{2} - A B + B^{2}) = A^{3} + B^{3}$	(Sum of Cubes)

The formula for the square or cube of trinomial (=polynomial with three terms) can also be obtained using the Square of a Sum and the Cube of a Sum formulas.

$(A + B + C)^{2} = A^{2} + B^{2} + C^{2} + 2 A B + 2 A C + 2 B C$
$(A + B + C)^{3} = A^{3} + B^{3} + C^{3} + 3 A^{2} B + 3 A B^{2} + 3 A^{2} C + 3 A C^{2} + 3 B^{2} C + 3 B C^{2}$

Examples

Example 1

Expand $(3 x - 5 y)^{2}$

Solution

Let $A = 3 x$ and $B = 5 y$ . Then

$\begin{aligned} (3 x - 5 y)^{2} & = (A - B)^{2} \\ = A^{2} - 2 A B + B^{2} \\ = (3 x)^{2} - 2 (3 x) (5 y) + (5 y)^{2} \\ = 9 x^{2} - 30 x y + 25 y^{2} \end{aligned}$

Example 2

Simplify $(x - y) (x + y) + y^{2}$

Solution

Because (Formula 6)

$(x - y) (x + y) = x^{2} - y^{2}$

we have

$\begin{aligned} (x - y) (x + y) + y^{2} & = x^{2} - y^{2} + y^{2} \\ = x^{2} . \end{aligned}$

Example 3

Simplify $(x^{3} - 2 \sqrt{y}) (x^{3} + 2 \sqrt{y})$

Solution

Let $A = x^{3}$ and $B = 2 \sqrt{y}$ . Then the above product can be written as $(A - B) (A + B)$ . Thus

$\begin{aligned} (x^{3} - 2 \sqrt{y}) (x^{3} + 2 \sqrt{y}) & = (A - B) (A + B) \\ = A^{2} - B^{2} \\ = {(x^{3})}^{2} - (2 \sqrt{y})^{2} \\ = x^{6} - 4 (\sqrt{y})^{2} \\ = x^{6} - 4 y \end{aligned}$

Note that the above equation has a meaning only when $y > 0$ .

Read more: Expansion of (A ± B)n

Collapse Expansion of (A ± B)n

Examples

Read more: Expansion of (A ± B)ⁿ

Collapse Expansion of (A ± B)ⁿ