Algebraic identities are equations where the value of the left-hand side of the equation is identically equal to the value of the right-hand side of the equation. Unlike algebraic expressions, algebraic identities satisfy all the values of the variables.
Let us consider an example to understand this better. Consider the equations: 5x - 3 = 12, 10x - 6 = 24, and x2 + 5x + 6 = 0. These equations satisfy only a unit value and do not work for any other values of the variables.
Now let us consider an equation x2 - 9 = (x + 3)(x - 3). These equations satisfy any values of x.
From this, we can clearly understand that an algebra identity has an expression that satisfies any values for the variables. The basic algebraic identities are primarily helpful to work out the numerous math problems.
All the standard Algebraic Identities are derived from the Binomial Theorem, which is given as:
Some Standard Algebraic Identities list are given below:
Identity I: (a + b)2 = a2 + 2ab + b2
Identity II: (a – b)2 = a2 – 2ab + b2
Identity III: a2 – b2= (a + b)(a – b)
Identity IV: (x + a)(x + b) = x2 + (a + b) x + ab
Identity V: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Identity VI: (a + b)3 = a3 + b3 + 3ab (a + b)
Identity VII: (a – b)3 = a3 – b3 – 3ab (a – b)
Identity VIII: a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
Example 1: Find the product of (x + 1)(x + 1) using standard algebraic identities.
Solution: (x + 1)(x + 1) can be written as (x + 1)2. Thus, it is of the form Identity I where a = x and b = 1. So we have,
(x + 1)2 = (x)2 + 2(x)(1) + (1)2 = x2 + 2x + 1
Example 2: Factorise (x4 – 1) using standard algebraic identities.
Solution: (x4 – 1) is of the form Identity III where a = x2 and b = 1. So we have,
(x4 – 1) = ((x2)2– 12) = (x2 + 1)(x2 – 1)
The factor (x2 – 1) can be further factorised using the same Identity III where a = x and b = 1. So,
(x4 – 1) = (x2 + 1)((x)2 –(1)2) = (x2 + 1)(x + 1)(x – 1)
Example 3: Factorise 16x2 + 4y2 + 9z2 – 16xy + 12yz – 24zx using standard algebraic identities.
Solution: 16x2 + 4y2 + 9z2– 16xy + 12yz – 24zx is of the form Identity V. So we have,
16x2 + 4y2 + 9z2 – 16xy + 12yz – 24zx = (4x)2 + (-2y)2 + (-3z)2 + 2(4x)(-2y) + 2(-2y)(-3z) + 2(-3z)(4x)= (4x – 2y – 3z)2 = (4x – 2y – 3z)(4x – 2y – 3z)
Example 4: Expand (3x – 4y)3 using standard algebraic identities.
Solution: (3x– 4y)3 is of the form Identity VII where a = 3x and b = 4y. So we have,
(3x – 4y)3 = (3x)3 – (4y)3– 3(3x)(4y)(3x – 4y) = 27x3 – 64y3 – 108x2y + 144xy2
Example 5: Factorize (x3 + 8y3 + 27z3 – 18xyz) using standard algebraic identities.
Solution: (x3 + 8y3 + 27z3 – 18xyz)is of the form Identity VIII where a = x, b = 2y and c = 3z. So we have,
(x3 + 8y3 + 27z3 – 18xyz) = (x)3 + (2y)3 + (3z)3 – 3(x)(2y)(3z)= (x + 2y + 3z)(x2 + 4y2 + 9z2 – 2xy – 6yz – 3zx)
The following proofs of algebraic identities will help us to visually understand each of the identities and better understand it. Let us look at the proofs of each of the basic algebraic identities.
Identity I: Proof of (a + b)2 = a2 + 2ab + b2
Identity II: Proof of (a − b)2 = a2 −2ab + b2
Identity III: Proof of (a + b)(a - b) = a2 - b2
Re-arranging the individual squares and rectangles, we get:
(a+b)(a−b)=a2−b2
Identity IV: Proof of (x + a)(x + b) = x2 + x(a + b) + ab
Proof of (x+a)(x+b)
Identity V: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
i.e., (a+b+c)2=a2+b2+c2+2ab+2bc+2ac
Hence, geometrically we proved the identity (a+b+c)2=a2+b2+c2+2ab+2bc+2ac
Identity VI: (a + b)3 = a3 + b3 + 3ab (a + b)
Lets draw a cube with side length (a+b) , hence we know that the volume of this cube would be equal to (a + b)3
Identity VII: (a – b)3 = a3 – b3 – 3ab (a – b)
For proof of identity (a-b)3 = a3-b3-3ab(a-b), let us consider a cube with side a and a small segment of side a be b as follows:
Split the cubes into small chucks to easily calculate volume as follows:
and splitting these cuboids to more simplified cuboids and cubes as follows:
Now, using the same volume remains constant before and after direction concepts.
a3=(a-b)3+b2(a-b)+(a-b)2b+(a-b)2b+b3+(a-b)2b+(a-b)b2+(a-b)b2
⇒ a3=(a-b)3+b(a-b)[b+a-b+a-b+a-b+b+b]+b3
⇒ a3=(a-b)3+b(a-b)[3a]+b3
⇒ a3=(a-b)3+3ab(a-b)+b3
Rearrenging this, we get (a-b)3 = a3-b3-3ab(a-b)
Hence, proved the identity (a-b)3 = a3-b3-3ab(a-b).
Identity VIII: (x + y + z)(x 2 + y 2 + z 2 – xy – yz – zx)
On expanding, we get the product as
x(x 2 + y 2 + z 2 – xy – yz – zx)
+ y(x 2 + y 2 + z 2 – xy – yz – zx)
+ z(x 2 + y 2 + z 2 – xy – yz – zx)
= x 3 + xy2 + xz 2 – x 2y – xyz – zx2 + x 2y + y 3 + yz2 – xy2 – y 2 z – xyz + x 2 z + y 2 z + z 3 – xyz – yz2 – xz2
= x 3 + y 3 + z 3 – 3xyz (On simplification)
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