Quick Reference
| # | Formula | Name |
|---|---|---|
| 1 | $(A+B)^{2}=A^{2}+2AB+B^{2}$ | Square of a Sum |
| 2 | $(A-B)^{2}=A^{2}-2AB+B^{2}$ | Square of a Difference |
| 3 | $(A+B)^{3}=A^{3}+3A^{2}B+3AB^{2}+B^{3}$ | Cube of a Sum |
| 4 | $(A-B)^{3}=A^{3}-3A^{2}B+3AB^{2}-B^{3}$ | Cube of a Difference |
| 6 | $(x+A)(x+B)=x^{2}+(A+B)x+AB$ | Product of Binomials with a Common Term |
| 7 | $(A+B)(A-B)=A^{2}-B^{2}$ | Difference of Squares |
| 8 | $(A-B)(A^{2}+AB+B^{2})=A^{3}-B^{3}$ | Difference of Cubes |
| 9 | $(A+B)(A^{2}-AB+B^{2})=A^{3}+B^{3}$ | Sum of Cubes |
The 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.
Here $A$ and $B$ represent real numbers, variables, or algebraic expressions.
Squares of Binomials
- $(A+B)^{2}=A^{2}+2AB+B^{2}$ (Square of a Sum)
- $(A-B)^{2}=A^{2}-2AB+B^{2}$ (Square of a Difference)
Cubes of Binomials
- $(A+B)^{3}=A^{3}+3A^{2}B+3AB^{2}+B^{3}$ (Cube of a Sum)
- $(A-B)^{3}=A^{3}-3A^{2}B+3AB^{2}-B^{3}$ (Cube of a Difference)
Note that the Square of a Difference formula can be obtained by replacing $B$ with $-B$ in the Square of a Sum formula. Similarly, replacing $B$ with $-B$ in the Cube of a Sum formula yields the Cube of a Difference formula.
Binomial Expansion
The formulas above handle the cases $n=2$ and $n=3$. For higher powers, the coefficients of the expansion of $(A+B)^{n}$ can be read directly from the Pascal-Newton-Khayyam triangle (commonly called Pascal's triangle). Each row of the triangle corresponds to one value of $n$, and the entries in that row are the coefficients of $A^{n}B^{0},\, A^{n-1}B^{1},\, \ldots,\, A^{0}B^{n}$ in order.
| $n$ | Coefficients of $(A+B)^n$ |
|---|---|
| 0 | $1$ |
| 1 | $1\quad 1$ |
| 2 | $1\quad 2\quad 1$ |
| 3 | $1\quad 3\quad 3\quad 1$ |
| 4 | $1\quad 4\quad 6\quad 4\quad 1$ |
| 5 | $1\quad 5\quad 10\quad 10\quad 5\quad 1$ |
| 6 | $1\quad 6\quad 15\quad 20\quad 15\quad 6\quad 1$ |
Each entry is the sum of the number directly above it and the number to the left of that one. For example, in row 4 the entry $6$ equals $3+3$ (the $3$ directly above it and the $3$ to its left in row 3), and each $4$ equals $1+3$ or $3+1$ by the same rule. The first and last entries of every row are always $1$. To expand $(A+B)^{n}$, read off row $n$, attach the corresponding powers of $A$ (descending from $n$ to $0$) and $B$ (ascending from $0$ to $n$), and write out the sum.
For example, using row $n=4$:
$ (A+B)^{4} = A^{4}+4A^{3}B+6A^{2}B^{2}+4AB^{3}+B^{4}. $And using row $n=5$:
$ (A+B)^{5} = A^{5}+5A^{4}B+10A^{3}B^{2}+10A^{2}B^{3}+5AB^{4}+B^{5}. $Further Study: The Binomial Coefficient Formula
The general formula for any power $n$
For large $n$, reading off the triangle row by row can be tedious. The entry in row $n$ at position $k$ (counting from $0$) is given by the **binomial coefficient** $ \binom{n}{k}=\frac{n!}{k!\,(n-k)!} \qquad \text{(read "$n$ choose $k$")} $ where $k!=1\times2\times\cdots\times k$ (with $0!=1$ by convention). This gives the general expansion: $ (A+B)^{n}=A^{n}+\binom{n}{1}A^{n-1}B+\binom{n}{2}A^{n-2}B^{2}+\cdots+\binom{n}{n-1}AB^{n-1}+B^{n} $ and $ \begin{aligned} (A-B)^{n}=A^{n}&-\binom{n}{1}A^{n-1}B+\cdots+(-1)^{k}\binom{n}{k}A^{n-k}B^{k}+\cdots \\ &+(-1)^{n-1}\binom{n}{n-1}AB^{n-1}+(-1)^{n}B^{n}. \end{aligned} $ You can verify that the binomial coefficients recover the triangle entries: $\binom{4}{0}=1$, $\binom{4}{1}=4$, $\binom{4}{2}=6$, $\binom{4}{3}=4$, $\binom{4}{4}=1$, which matches row $n=4$ above.Other Product Formulas
- $(x+A)(x+B)=x^{2}+(A+B)x+AB$ (Product of Binomials with a Common Term)
- $(A+B)(A-B)=A^{2}-B^{2}$ (Difference of Squares)
- $(A-B)(A^{2}+AB+B^{2})=A^{3}-B^{3}$ (Difference of Cubes)
- $(A+B)(A^{2}-AB+B^{2})=A^{3}+B^{3}$ (Sum of Cubes)
Trinomial Formulas
The square and cube of a trinomial $A+B+C$ can be derived from the binomial formulas by treating two of the three terms as a single unit. Set $D=A+B$, so that $A+B+C=D+C$, and then apply the binomial formulas to the pair $(D,C)$.
Square of a trinomial. Apply the Square of a Sum formula to $(D+C)^{2}$:
$ \begin{aligned} (A+B+C)^{2}&=(D+C)^{2}=D^{2}+2DC+C^{2} \\ &=(A+B)^{2}+2(A+B)C+C^{2} \\ &=(A^{2}+2AB+B^{2})+(2AC+2BC)+C^{2} \\ &=A^{2}+B^{2}+C^{2}+2AB+2AC+2BC. \end{aligned} $In other words, the square of a trinomial equals the sum of the squares of each term plus twice the product of every pair of distinct terms: $A^{2}$, $B^{2}$, $C^{2}$ from squaring each term, and $2AB$, $2AC$, $2BC$ from doubling each pairwise product.
Derivation of the cube of a trinomial
Apply Formula 3 to $(D+C)^{3}$, where $D=A+B$: $ (A+B+C)^{3}=(D+C)^{3}=D^{3}+3D^{2}C+3DC^{2}+C^{3}. $ Expand each term using the binomial formulas: $ \begin{aligned} D^{3}&=(A+B)^{3}=A^{3}+3A^{2}B+3AB^{2}+B^{3}, \\ 3D^{2}C&=3(A+B)^{2}C=3(A^{2}+2AB+B^{2})C=3A^{2}C+6ABC+3B^{2}C, \\ 3DC^{2}&=3(A+B)C^{2}=3AC^{2}+3BC^{2}, \\ C^{3}&=C^{3}. \end{aligned} $ Adding all terms: $ (A+B+C)^{3}=A^{3}+B^{3}+C^{3}+3A^{2}B+3AB^{2}+3A^{2}C+3AC^{2}+3B^{2}C+3BC^{2}+6ABC. $ In other words, the cube of a trinomial equals the sum of the cubes of each term ($A^{3}$, $B^{3}$, $C^{3}$), plus three times the square of each term multiplied by each of the other two terms ($3A^{2}B$, $3AB^{2}$, $3A^{2}C$, $3AC^{2}$, $3B^{2}C$, $3BC^{2}$), plus six times the product of all three terms ($6ABC$).Common Mistake: Squaring a Binomial
Warning: $(A+B)^{2} \neq A^{2}+B^{2}$.
The middle term $2AB$ is always present and must not be omitted. For example,
$(x+3)^{2} = x^{2}+6x+9$, not $x^{2}+9$.
Similarly, $(A-B)^{2} \neq A^{2}-B^{2}$; the correct expansion is $A^{2}-2AB+B^{2}$.
Worked Examples
Example 1. Expand $(3x-5y)^{2}$.
Solution
Let $A=3x$ and $B=5y$. Using Formula 2: $ \begin{aligned} (3x-5y)^{2}&=(A-B)^{2} \\ &=A^{2}-2AB+B^{2} \\ &=(3x)^{2}-2(3x)(5y)+(5y)^{2} \\ &=9x^{2}-30xy+25y^{2} \end{aligned} $Example 2. Simplify $(x-y)(x+y)+y^{2}$.
Solution
Using Formula 7 (Difference of Squares), $(x-y)(x+y)=x^{2}-y^{2}$. Therefore: $ \begin{aligned} (x-y)(x+y)+y^{2}&=x^{2}-y^{2}+y^{2} \\ &=x^{2} \end{aligned} $Example 3. Find $(x^{3}-2\sqrt{y})(x^{3}+2\sqrt{y})$.
Solution
Let $A=x^{3}$ and $B=2\sqrt{y}$. The product has the form $(A-B)(A+B)$, so by Formula 7: $ \begin{aligned} (x^{3}-2\sqrt{y})(x^{3}+2\sqrt{y})&=(A-B)(A+B) \\ &=A^{2}-B^{2} \\ &=\left(x^{3}\right)^{2}-\left(2\sqrt{y}\right)^{2} \\ &=x^{6}-4y \end{aligned} $ Note that the above equation has a meaning only when $y\geq 0$.Frequently Asked Questions