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.

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

Binomial expansion: 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}+{n \choose 1}A^{n-1}B+{n \choose 2}A^{n-2}B+\cdots+{n \choose n-1}AB+B^{n} $

and

$ \begin{aligned} (A-B)^{n}=A^{n}-&{n \choose 1}A^{n-1}B+\cdots+(-1)^{k}{n \choose k}A^{n-k}B^{k}+\cdots &+(-1)^{n-1}{n \choose n-1}AB^{n-1}+(-1)^{n}B^{n} \end{aligned} $

where

$ {n \choose k}=\frac{n!}{k!(n-k)!} \quad (read "$n$ choose $k$") $

with $k!=1\times2\times3\times\cdots\times(k-1)\times k.$

For example,

$\begin{aligned} (x+y)^{4} & =x^{4}+{4 \choose 1}x^{3}y+{4 \choose 2}x^{2}y^{2}+{4 \choose 3}x^{3}y+y^{4} & =x^{4}+\frac{4!}{1!\times3!}x^{3}y+\frac{4!}{2!\times2!}x^{2}y^{2}+\frac{4!}{3!\times1!}x^{3}y+y^{4} & =x^{4}+4x^{3}y+6x^{2}y^{2}+4xy^{3}+y^{4} \end{aligned}$

$(x+A)(x+B)=x^{2}+(A+B)x+AB$ \hfill (Product of two Binomials having a Common Term)
$(A-B)(A+B)=A^{2}-B^{2}$\hfill (Product of Sum and Difference)
$(A-B)(A^{2}+AB+B^{2})=A^{3}-B^{3}$\hfill (Difference of Cubes)
$(A+B)(A^{2}-AB+B^{2})=A^{3}+B^{3}$ \hfill (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.

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

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

Solution

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

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

Solution

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

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

Solution

Let $A=x^{3}$ and $B=2\sqrt{y}$. Then the above product can be written as $(A-B)(A+B)$. Thus $\begin{aligned} (x^{3}-2\sqrt{y})(x^{3}+2\sqrt{y}) & =(A-B)(A+B) & =A^{2}-B^{2} & =\left(x^{3}\right)^{2}-(2\sqrt{y})^{2} & =x^{6}-4(\sqrt{y})^{2} & =x^{6}-4y \end{aligned}$ Note that the above equation has a meaning only when $y>0$.

Special Product Formulas

Special Product Formulas

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