From the multiplication rules for sums and polynomials can easily obtain the following seven special factors.

They should know by heart, as they are used in all problems in mathematics.

[1] *( a + b )**² = a² + 2ab + b² ,*

[2] ** ( a **–

*b )***–**

*² = a²*

*2ab + b*

*² ,*[3] ** ( a + b ) ( a **–

*b ) = a***–**

*²*

*b*

*²,*[4] *( a + b )**³**= a**³ + 3a² b + 3ab² + b³ ,*

[5] ** ( a **–

*b )*

*³*

*= a***–**

*³*

*3a***–**

*² b + 3ab²*

*b*

*³ ,*[6] *( a + b )( a*** ² **–

*ab + b*

*² ) = a³ + b³ ,*[7]** ( a **–

*b )( a*

*² +*

*ab + b***–**

*² ) = a³*

*b*

*³*Let’s look some examples from SAT tests :

**Example 1**

If (*m *+ *n*)^{2} = 18 and *mn *= 4, then what is the value of m^{2} + *n ^{2}*?

Following the formula [1], and using first condition, we will get:** **

* m ^{2}+2mn+n^{2 }= 18,*

* *So, if we will use second condition, we will get :

*m ^{2}+8+n^{2 }= 18, so *m

^{2}+

*n*

^{2 }= 10.**Example 2**

If *(a-b) = -4*, what is the value of *a ^{2}-2ab + b^{2}* ?

Following the formula [2], we will get:

* a ^{2 }– 2ab + b^{2 }*=

*( a*–

*b )² = (-4)*

^{2 }= 16**Example 3**

If *ab + 1/(ab) = 4*, what is the value *(ab) ^{2 }+1/(ab)^{2}*

From *ab + 1/(ab) = 4* and formula [1], we will get*:*

* (ab + 1/(ab) ) ^{2} = (ab)^{2 }+2*

*⨯ ab*

*⨯ 1/(ab) + 1/(ab)*

^{2}Reducing the fraction, we will get:

* (ab) ^{2 }+2+1/(ab)^{2 }= 16*, so

*(ab)*

^{2 }+1/(ab)^{2 }= 14