Algebraicidentities :
Here we are going to see the list of identities which are being used to solve all kind of problems in the topic Algebra in math.
Algebraic identities is an equality which remains true regardless of the value of any variable which appear within it.
Formula is the short cut of doing problems without doing the complete steps. It is very long process of doing any problem without formula. Let us consider the following example to understand the purpose of formulas in math.
Formulas |
Examples & Practice questions |
(1) (a + b)² = a² + 2 ab + b² (2) (a + b)² = (a - b)² + 4 ab | |
(3) (a - b)² = a² - 2 ab + b² (4) (a - b)² = (a + b)² - 4 ab | |
(5) a² - b² = (a + b) (a - b) | |
(6) (x+a) (x+b) =x² + (a + b) x+a b | |
(7) (a+b)³=a³+3a²b+3ab²+b³ (8) (a+b)³=a³+b³+3ab(a+b) | |
(9) (a-b)³=a³-3a²b+3ab²-b³ (10) (a-b)³=a³-b³-3ab(a-b) | |
(11) a³+b³ = (a+b)(a²-ab+b²) (12) a³+b³=(a+b)³-3 ab(a + b) | |
(13) a³-b³= (a-b)(a²+ab+ b²) (14) a³-b³=(a-b)³ +3ab(a-b) |
(15) (a+b+c)²= a²+b²+c² +2ab+2bc+2ca (16) (a+b-c)²=a²+b²+c² +2ab-2bc-2ca (17) (a-b+c)²= a²+b²+c²-2ab-2bc+2ca (18) (a-b-c)²= a²+b²+c²-2ab+2bc-2ca (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] |
(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
We can remember the expansion of the identities like (a+b)² (a+b+c)², (a+b+c)³. In the above identities, if one or more terms is negative, how can we remember the expansion?
This question has been answered in the following two cases.
Case 1 :
For example, let us consider the identity of (a + b + c)²
We can easily remember the expansion of (a + b + c)².
If "c" is negative, we will have (a + b - c)²
How can we remember the expansion of (a + b - c)² ?
It is very simple.
In the terms of the expansion, a², b², c², ab, bc, ca, let us consider the terms in which we find "c"
They are c², bc, ca .
Even if we take negative sign for "c", 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" which is negative.
Finally, we have
(a + b - c)²= a² + b² + c² + 2ab - 2bc - 2ca
Case 2 :
In (a+b+c)², if both "b" and "c" are negative, we will have (a - b - c)²
How can we remember the expansion of (a - b - c)² ?
It is very simple.
In the terms of the expansion, a², b², c², ab, bc, ca, let us consider the terms in which we find "b" and "c"
They are b², c², ab, bc, ca.
Even if we take negative sign for "b" and "c", the sign of b² and c² will be positive. Because they even power "2".
The terms "ab" and "ca" will be negative.
Because, in "ab", "a" is multiplied by "b" which is negative.
Because, in "ca", "a" is multiplied by "c" which is negative.
The term "bc" will be positive.
Because, in "bc", both "b" and "c" are negative. (negative x negative = positive)
Finally, we have
(a - b - c)²= a² + b² + c² - 2ab + 2bc - 2ca
In the same we can get idea to remember the the expansions of (a + b - c)³, (a - b + c)³ (a - b - c)³
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