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.
- \((A+B)^{2}=A^{2}+2AB+B^{2}\) (Square of a Sum)
- \((A-B)^{2}=A^{2}-2AB+B^{2}\) (Square of a Difference)
- \((A+B)^{3}=A^{3}+3A^{2}B+3AB^{2}+B^{3}\) (Cube of a Sum)
- \((A-B)^{3}=A^{3}-3A^{2}B+3AB^{2}-B^{3}\) (Cube of a Difference)
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 \[\begin{align} (A-B)^{n}=A^{n}-&{n \choose 1}A^{n-1}B+\cdots+(-1)^{k}{n \choose k}A^{n-k}B^{k}+\cdots\\ &+(-1)^{n-1}{n \choose n-1}AB^{n-1}+(-1)^{n}B^{n} \end{align}\] where \[{n \choose k}=\frac{n!}{k!(n-k)!}\tag{read ``$n$ choose $k$"}\] with \(k!=1\times2\times3\times\cdots\times(k-1)\times k.\)
For example, \[\begin{align} (x+y)^{4} & =x^{4}+{4 \choose 1}x^{3}y+{4 \choose 2}x^{2}y^{2}+{4 \choose 3}x^{3}y+y^{4}\\ & =x^{4}+\frac{4!}{1!\times3!}x^{3}y+\frac{4!}{2!\times2!}x^{2}y^{2}+\frac{4!}{3!\times1!}x^{3}y+y^{4}\\ & =x^{4}+4x^{3}y+6x^{2}y^{2}+4xy^{3}+y^{4} \end{align}\]
- \((x+A)(x+B)=x^{2}+(A+B)x+AB\) (Product of two Binomials having a Common Term)
- \((A-B)(A+B)=A^{2}-B^{2}\) (Product of Sum and Difference)
- \((A-B)(A^{2}+AB+B^{2})=A^{3}-B^{3}\) (Difference of Cubes)
- \((A+B)(A^{2}-AB+B^{2})=A^{3}+B^{3}\) (Sum of Cubes)
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.
- \((A+B+C)^{2}=A^{2}+B^{2}+C^{2}+2AB+2AC+2BC\)
- \((A+B+C)^{3}=A^{3}+B^{3}+C^{3}+3A^{2}B+3AB^{2}+3A^{2}C+3AC^{2}+3B^{2}C+3BC^{2}\)
Example 1.18 . Expand \((3x-5y)^{2}\) .
Solution
Let \(A=3x\) and \(B=5y\) . Then \[\begin{align} (3x-5y)^{2} & =(A-B)^{2}\\ & =A^{2}-2AB+B^{2}\\ & =(3x)^{2}-2(3x)(5y)+(5y)^{2}\\ & =9x^{2}-30xy+25y^{2} \end{align}\]
Example 1.19 . Simplify \((x-y)(x+y)+y^{2}\) .
Solution
Because \[(x-y)(x+y)=x^{2}-y^{2}\] we have \[\begin{align} (x-y)(x+y)+y^{2} & =x^{2}-y^{2}+y^{2}\\ & =x^{2}. \end{align}\]
Example 1.20 . 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 \[\begin{align} (x^{3}-2\sqrt{y})(x^{3}+2\sqrt{y}) & =(A-B)(A+B)\\ & =A^{2}-B^{2}\\ & =\left(x^{3}\right)^{2}-(2\sqrt{y})^{2}\\ & =x^{6}-4(\sqrt{y})^{2}\\ & =x^{6}-4y \end{align}\] Note that the above equation has a meaning only when \(y>0\) .