Algebraic Identities, the topic with which most of the students are feared of solving mathematical problems. But I hope you won’t be feared anymore after going through this article completely. So, without wasting more time let’s start this topic, though considered difficult by you, yet easy.
What is an Algebraic Identity?
Algebraic Identity is an equality which holds true for any real value given to the variables in the given equation i.e. for any value of the variables, the left-hand side(L.H.S) will be equal to the right-hand side(R.H.S).
Eg: (a + 1)(a +2) = a2+3a+2
=> In the given equality, let’s put a = 10 in both sides.
LHS = (a + 1) (a + 2) = (10 + 1) (10 + 2) = 11 × 12 = 132
RHS = a2 + 3a + 2 = 102 + 3 × 10 + 2 = 100 + 30 + 2 = 132
For a = 10, L.H.S = R.H.S
Similarly, you can check for any value of a and you will see L.H.S is always to equal to R.H.S, which makes this equality an identity.
An equality true for every value of the variable in it is called an identity.
Is every equation which holds true is an identity?
The answer is “NO”. The reason is that an equation is true for only certain values of the variable in it. It is not true for all values of the variable.
For eg: a2 + 3a + 2 = 132.
=> It is true for a =10 but not for a = 0 or any other value. So, it is an equation but not an identity.
Now, when you know about identities, shouldn’t we enhance our knowledge by learning some common algebraic identities?
Standard Algebraic Identities:
1. (a+b)2 = a2 + 2ab + b2
2. (a-b)2 = a2 – 2ab + b2
3. (a+b) (a-b) = a2 -b2
4. (x + a)(x + b) = x2 + (a + b) x + ab
5. (x + a)(x – b) = x2 + (a – b) x – ab
6. (x – a)(x + b) = x2 + (b – a) x – ab
7. (x – a)(x – b) = x2 – (a + b)x + ab
8. (a + b)3 = a3 + b3 +3ab (a + b)
9. (a – b)3 = a3 – b3 – 3ab (a – b)
10. (x + y + z)2 = x2 + y2 + z2 + 2xy +2yz + 2xz
11. (x + y – z)2 = x2 + y2 + z2 + 2xy – 2yz – 2xz
12. (x – y + z)2 = x2 + y2 + z2 – 2xy – 2yz + 2xz
13. (x – y – z)2 = x2 + y2 + z2 – 2xy + 2yz – 2xz
14. x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz -xz)
15. x2 + y2 = (1/2)[(x+y)2 + (x-y)2]
16. (x+a) (x+b) (x+c) = x3+ (a + b + c ) x2 + (ab + bc +ca) x + abc
17. x3 + y3 = (x + y) ( x2 – xy + y2)
18. x3 – y3 = (x – y) ( x2 + xy + y2)
19. x2 + y2 + z2 -xy -yz -zx = (1/2)[(x-y)2 + (y -z)2 + (z -x)2]
This list of standard algebraic identities appears to be intimidating but need not worry we are going to solve questions based on these identities as well which will make these much easier for you.
1. (a+b)2 = a2 + 2ab + b2
Let’s prove this identity:
(a+b)2 = ( a + b) ( a + b)
= a ( a + b) + b ( a + b) (Multiplying every term with each other)
= a2 + ab + ba + b2
= a2 + 2ab + b2
Further, let a = 1 and b= 2
(1+2)2 = 12 + 2*1*2 + 22
9 = 9
L.H.S = R.H.S
Hence Proved
2. (a – b)2 = a2 – 2ab + b2
= (a – b) (a – b) = a (a – b) – b (a – b)
= a2 – ab – ba + b2
= a2 – 2ab + b2
Let a = 1 and b = 2
(1-2)2 = 12– 2*1*2 + 22
1 = 1
L.H.S = R.H.S
Hence Proved
3. (a+b) (a-b) = a2 -b2
= (a + b) (a – b) = a (a – b) + b (a – b)
= a2 – ab + ba – b2
= a2 – b2
Let a = 1 and b = 2
(1+2)(1-2) = 12– 22
-3 = -3
L.H.S = R.H.S
Hence Proved
In a similar way, you can prove every identity and you will see that these all satisfy the condition to be an identity. But I doubt this won’t fulfill the purpose of making you learn these identities so that you can use them while solving problems.
How to remember algebraic identities?
One you can do this by learning through memorization, by rote learning which is followed by repetition. You can repeat all the identities again and again orally or by writing until you cram them effectively but I wonder you will end committing mistake most of the time by following this way.
Another way which I think can prove more effective is by practicing problems after once you go through every identity.
Sample Problems:
Example 1: Find the product of (x + 2)(x + 2) using standard algebraic identities.
Solution: (x + 2)(x + 2) can be written as (x + 2)2.
Thus, after comparing with standard identity here a = x and b = 2. So we have,
(x + 2)2 = (x)2 + 2(x)(2) + (2)2 = x2 + 4x + 4.
Example 2: Expand ( x – 2y)2
Solution : Use identity ( a -b )2 = a2 – 2ab + b2 .
After comparing, a = x, b = 2y. So, we have ( x)2 – 2 * x * 2y + (2y) = x2 – 4xy + 4y2
Example 3: Factorise 16x2 + 4y2 + 9z2 – 16xy + 12yz – 24zx using standard algebraic identities.
Solution: 16x2 + 4y2 + 9z2– 16xy + 12yz – 24zx is of the form (x – y – z)2 = x2 + y2 + z2 – 2xy + 2yz – 2xz.
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 (2x – 4y)3 using standard algebraic identities.
Solution: (2x– 4y)3 is of the form (a – b)3 = a3 – b3 – 3ab (a – b) where a = 2x and b = 4y.
So we have,
(2x – 4y)3 = (2x)3 – (4y)3– 3(2x)(4y)(2x – 4y) = 8x3 – 64y3 – 48x2y + 96xy2
Example 5: Expand 8a3 – 27b3 using standard algebraic identities.
Solution: 8a3– 27b3 is of the form x3 – y3 = (x – y) ( x2 + xy + y2) where x = 2a and y = 3b.
So we have,
8a3– 27b3 = (2a – 3b) ( 4a2 + 6ab + 9b2).
These were some of the basic sample problems for your better understanding.
You should also try to solve similar problems by yourself now. So, that you can have a good grip on algebraic identities.
Here are some homework problems which you must solve by yourself.
Q.1 Expand x3+1 using standard algebraic identities.
Q.2 Expand (5x + 3y + 2z )² using standard algebraic identities.
Q.3 Expand ((2x)³ – y³) using standard algebraic identities.
Q.4 If a + b + c = 10 , ab + bc + ca = 20 find a² + b² + c².
Try these simple problems and if you have any doubt, feel free to ask.
This was all about algebraic identities according to the class 9th syllabus. I hope you understood the concepts well. Stay tuned for more such articles.
