Algebra Formulas

ALGEBRA FORMULAS

Algebra formulas are equalities which remain true regardless of the values of any variables which appear within it.

In our website, we have provided two calculators for algebra formulas.

One is to find the expansion for (a + b)ⁿ and other one is to find the expansion for (a - b)ⁿ.

Please click the below links to get the linear regression needed.

Expansion Calculator for (a + b)ⁿ

Expansion Calculator for (a - b)ⁿ

If you would like to have problems on algebraic identities, please click the link given below.

Worksheet on Algebraic Identities

Proving Algebraic Identity Expansion Geometrically

In this section, you will learn how to prove the expansions of algebraic identities geometrically.

Let us consider algebraic identity and its expansion given below.

(a + b)² = a² + 2ab + b²

We can prove the the expansion of (a + b)² using the area of a square as shown below.

Algebraic Formulas

In this section, we are going to see the list of formulas which are being used to solve all kind of problems in the Algebra.

(a + b)² = a² + 2ab + b² (a + b)² = (a - b)² + 4ab	Examples
(a - b)² = a² - 2ab + b² (a - b)² = (a + b)² - 4ab	Examples
a² - b² = (a + b)(a - b)	Examples
(x + a)(x + b) = x² + (a + b)x + ab	Examples
(a + b)³= a³+ 3a²b + 3ab²+ b³ (a + b)³= a³+ 3ab(a + b)+ b³	Examples
(a - b)³= a³- 3a²b + 3ab²- b³ (a - b)³= a³- 3ab(a - b)- b³	Examples
a³+ b³ = (a + b)(a²- ab + b²) a³- b³ = (a - b)(a²+ ab + b²)	Examples
a³+ b³ = (a + b)³ - 3ab(a + b) a³- b³ = (a - b)³ + 3ab(a - b)	Examples

(a + b + c)²= a²+ b²+ c² + 2ab + 2bc + 2ac

(a + b - c)²= a²+ b²+ c² + 2ab - 2bc - 2ac

(a - b + c)²= a²+ b²+ c² - 2ab - 2bc + 2ac

(a - b - c)²= a²+ b²+ c² - 2ab + 2bc - 2ac

a² + b² = (a + b)² - 2ab

a² + b² = (a - b)² + 2ab

a² + b² = 1/2 ⋅ [(a + b)² + (a - b)²]

ab = 1/4 ⋅ [(a + b)² - (a - b)²]

(a + b + c)³ = a³ + b³ + c³ + 3a²b + 3a²c + 3ab² + 3b²c + 3ac³² + 3bc² + 6abc

(a + b - c)³ = a³ + b³ - c³ + 3a²b - 3a²c + 3ab² - 3b²c + 3ac² + 3bc² - 6abc

(a - b + c)³ = a³ - b³ + c³ - 3a²b + 3a²c + 3ab² + 3b²c + 3ac² - 3bc² - 6abc

(a - b - c)³ = a³ - b³ - c³ - 3a²b - 3a²c + 3ab² - 3b²c + 3ac² - 3bc² + 6abc

(19) a² + b² = (a + b)² - 2ab

(20) a² + b² = (a - b)² + 2ab

(21) a² + b²=½ [(a+b)²-(a-b)²]

(22) ab = ¼[(a+b)²- (a - b)²]

Examples

Practice Questions

(23) (a + b + c)³ = a³ + b³ + c³ + 3a²b + 3a²c + 3ab² + 3b²c + 3ac² + 3bc² + 6abc

(24) (a + b - c)³ = a³ + b³ - c³ + 3a²b - 3a²c + 3ab² - 3b²c + 3ac² + 3bc² - 6abc

(25) (a - b + c)³ = a³ - b³ + c³ - 3a²b + 3a²c + 3ab² + 3b²c + 3ac² - 3bc² - 6abc

(26) (a - b - c)³ = a³ - b³ - c³ - 3a²b - 3a²c + 3ab² - 3b²c + 3ac² - 3bc² + 6abc

How to remember algebra formulas with negative sign?

We can remember algebraic formula expansions like

(a + b)², (a + b + c)², (a + b + c)³

In the above formulas, if one or more terms is negative, how can we remember the expansion ?

This question has been answered in the following three cases.

Case 1 :

We can easily remember the expansion of (a + b + c)².

If c is negative, then we will have

(a + b - c)²

How can we remember the expansion of (a + b - c)² ?

It is very simple.

Let us consider the expansion of (a + b + c)²_.

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

In the terms of the expansion above, consider the terms in which we find "c".

They are c², bc, ca.

Even if we take negative sign for "c" in c², the sign of c² will be positive. Because it has even power 2.

The terms bc, ca will be negative. Because both "b" and "a" are multiplied by "c" that is negative.

Finally, we have

(a + b - c)² = a² + b² + c² + 2ab - 2bc - 2ca

Case 2 :

In (a + b + c)², if "b" is negative, then we will have

(a - b + c)²

How can we remember the expansion of (a - b + c)² ?

It is very simple.

Let us consider the expansion of (a + b + c)²_.

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

In the terms of the expansion above, consider the terms in which we find "b".

They are b², ab, bc.

Even if we take negative sign for "b" in b², the sign of b² will be positive. Because it has even power 2.

The terms ab, bc will be negative. Because both "a" and "c" are multiplied by "b" that is negative.

Finally, we have

(a - b + c)² = a² + b² + c² - 2ab - 2bc + 2ca

Case 3 :

In (a + b + c)², if both "b" and "c" are negative, then we will have

(a - b - c)²

How can we remember the expansion of (a - b - c)² ?

It is very simple.

Let us consider the expansion of (a + b + c)²_.

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

In the terms of the expansion above, consider the terms in which we find "b" and "c".

They are b², c², ab, bc, ac.

Even if we take negative sign for "b" in b² and negative sign for "c" in c², the sign of both b²and c² will be positive. Because they have even power 2.

The terms "ab" and "ca" will be negative.

Because, in "ab", "a" is multiplied by "b" that is negative.

Because, in "ca", "a" is multiplied by "c" that is negative.

The term "bc" will be positive.

Because, in "bc", both "b" and "c" are negative.

That is,

negative ⋅ negative = positive

Finally, we have

(a - b - c)² = a² + b² + c² - 2ab + 2bc - 2ca

In the same way, we can get idea to remember the the expansions of

(a + b - c)³, (a - b + c)³, (a - b - c)³