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.
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(a + b)2 = a2 + 2ab + b2 (a + b)2 = (a - b)2 + 4ab | |
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(a - b)2 = a2 - 2ab + b2 (a - b)2 = (a + b)2 - 4ab | |
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a2 - b2 = (a + b)(a - b) | |
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(x + a)(x + b) = x2 + (a + b)x + ab | |
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(a + b)3 = a3 + 3a2b + 3ab2 + b3 (a + b)3 = a3 + 3ab(a + b) + b3 | |
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(a - b)3 = a3 - 3a2b + 3ab2 - b3 (a - b)3 = a3 - 3ab(a - b) - b3 | |
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a3 + b3 = (a + b)(a2 - ab + b2) a3 - b3 = (a - b)(a2 + ab + b2) | |
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a3 + b3 = (a + b)3 - 3ab(a + b) a3 - b3 = (a - b)3 + 3ab(a - b) |
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(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] |
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 b2 and 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
Apart from the stuff given above, if you would like to have problems on algebraic identities, please click the link given below.
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