How comes (a²+b²)= a²+b²+2ab ? Prove.
Ans
We can consider a Square area which is divided into four area inside.Having sides are (a+b)
Proving Algebraic Identities (Easy Notes)
The Main Idea
An identity is true for every value of a and b. To prove one, we multiply it out step by step. Then we check it with numbers.
Two rules you need:
- a^2 means a × a
- ab = ba (order doesn’t matter in multiplication)
1. (a + b)^2 = a^2 + 2ab + b^2
Proof: (a + b)^2 means (a + b) × (a + b).
Multiply every term of the first bracket with every term of the second:
- a × a = a^2
- a × b = ab
- b × a = ab
- b × b = b^2
Add them: a^2 + ab + ab + b^2 = a^2 + 2ab + b^2 ✔
Number check: a = 3, b = 2
- Left side: (3 + 2)^2 = 5^2 = 25
- Right side: 9 + 2(3)(2) + 4 = 9 + 12 + 4 = 25 ✔
Picture proof (area): Take a square with side (a + b). Cut it into 4 parts:
- one square a × a = a^2
- one square b × b = b^2
- two rectangles a × b = ab each
Total area = a^2 + b^2 + ab + ab = a^2 + 2ab + b^2
2. (a − b)^2 = a^2 − 2ab + b^2
Proof: (a − b)(a − b)
- a × a = a^2
- a × (−b) = −ab
- (−b) × a = −ab
- (−b) × (−b) = +b^2 (minus × minus = plus)
Add: a^2 − ab − ab + b^2 = a^2 − 2ab + b^2 ✔
Number check: a = 5, b = 2
- Left: (5 − 2)^2 = 9
- Right: 25 − 20 + 4 = 9 ✔
3. (a + b)(a − b) = a^2 − b^2
Proof:
- a × a = a^2
- a × (−b) = −ab
- b × a = +ab
- b × (−b) = −b^2
The middle terms −ab and +ab cancel each other. What is left: a^2 − b^2 ✔
Number check: a = 7, b = 3
- Left: (10)(4) = 40
- Right: 49 − 9 = 40 ✔
Use: Quick multiplication. 52 × 48 = (50 + 2)(50 − 2) = 2500 − 4 = 2496
4. (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
Proof: (a + b)^3 = (a + b) × (a + b)^2
We already know (a + b)^2 = a^2 + 2ab + b^2. So multiply:
(a + b)(a^2 + 2ab + b^2)
Step 1: multiply by a:
a^3 + 2a^2b + ab^2
Step 2: multiply by b:
a^2b + 2ab^2 + b^3
Step 3: add them, combining like terms:
- a^3
- 2a^2b + a^2b = 3a^2b
- ab^2 + 2ab^2 = 3ab^2
- b^3
Result: a^3 + 3a^2b + 3ab^2 + b^3 ✔
Number check: a = 2, b = 1
- Left: 3^3 = 27
- Right: 8 + 3(4)(1) + 3(2)(1) + 1 = 8 + 12 + 6 + 1 = 27 ✔
Picture proof (volume): A cube of side (a + b) splits into:
- 1 big cube a^3
- 1 small cube b^3
- 3 boxes of size a × a × b
- 3 boxes of size a × b × b
5. (a − b)^3 = a^3 − 3a^2b + 3ab^2 − b^3
Proof: Same method. (a − b)(a^2 − 2ab + b^2)
- Multiply by a: a^3 − 2a^2b + ab^2
- Multiply by (−b): −a^2b + 2ab^2 − b^3
Add: a^3 − 3a^2b + 3ab^2 − b^3 ✔
Number check: a = 3, b = 1
- Left: 2^3 = 8
- Right: 27 − 27 + 9 − 1 = 8 ✔
6. a^3 + b^3 = (a + b)(a^2 − ab + b^2)
Proof: Expand the right side.
(a + b)(a^2 − ab + b^2)
- a × (a^2 − ab + b^2) = a^3 − a^2b + ab^2
- b × (a^2 − ab + b^2) = a^2b − ab^2 + b^3
Add: a^3 + b^3 (the other terms cancel) ✔
7. a^3 − b^3 = (a − b)(a^2 + ab + b^2)
Proof:
- a × (a^2 + ab + b^2) = a^3 + a^2b + ab^2
- (−b) × (a^2 + ab + b^2) = −a^2b − ab^2 − b^3
Add: a^3 − b^3 ✔
Number check (for 6): a = 2, b = 1
- Left: 8 + 1 = 9
- Right: (3)(4 − 2 + 1) = 3 × 3 = 9 ✔
| Identity | Result |
|---|---|
| (a + b)^2 | a^2 + 2ab + b^2 |
| (a − b)^2 | a^2 − 2ab + b^2 |
| (a + b)(a − b) | a^2 − b^2 |
| (a + b)^3 | a^3 + 3a^2b + 3ab^2 + b^3 |
| (a − b)^3 | a^3 − 3a^2b + 3ab^2 − b^3 |
| a^3 + b^3 | (a + b)(a^2 − ab + b^2) |
| a^3 − b^3 | (a − b)(a^2 + ab + b^2) |
8. (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca
Proof: Multiply (a + b + c)(a + b + c). Every term meets every other term:
- Squares: a^2, b^2, c^2
- Pairs: ab appears twice, bc twice, ca twice
- Left: 6^2 = 36
- Right: 1 + 4 + 9 + 2(2) + 2(6) + 2(3) = 14 + 4 + 12 + 6 = 36 ✔
Result: a^2 + b^2 + c^2 + 2ab + 2bc + 2ca ✔
Check: a = 1, b = 2, c = 3
9.(x + a)(x + b) = x^2 + (a + b)x + ab
Proof:
- x × x = x^2
- x × b = bx
- a × x = ax
- a × b = ab
- Left: (7)(8) = 56
- Right: 25 + 5(5) + 6 = 25 + 25 + 6 = 56 ✔
Check: x = 5, a = 2, b = 3
Use: (x + 3)(x + 4) = x^2 + 7x + 12 (add the numbers, multiply the numbers).
10. (a + b)^2 + (a − b)^2 = 2(a^2 + b^2)
Proof:
- (a + b)^2 = a^2 + 2ab + b^2
- (a − b)^2 = a^2 − 2ab + b^2
Add them: the +2ab and −2ab cancel. Left: 2a^2 + 2b^2 = 2(a^2 + b^2) ✔
Check: a = 3, b = 1
- Left: 16 + 4 = 20
- Right: 2(9 + 1) = 20 ✔
11. (a + b)^2 − (a − b)^2 = 4ab
Proof: Subtract the same two expansions.
(a^2 + 2ab + b^2) − (a^2 − 2ab + b^2)
The a^2 and b^2 cancel. Left: 2ab + 2ab = 4ab ✔
Check: a = 4, b = 2
- Left: 36 − 4 = 32
- Right: 4(4)(2) = 32 ✔
Use: 51^2 − 49^2 = 4(25)(1)... more simply, a = 50, b = 1 gives 4 × 50 × 1 = 200. (Also (51 − 49)(51 + 49) = 2 × 100 = 200 ✔)
12. (a + b)^2 = (a − b)^2 + 4ab
Proof: From identity 4 above, move (a − b)^2 to the other side. ✔
Check: a = 5, b = 2
- Left: 49
- Right: 9 + 40 = 49 ✔
Use: If a − b = 3 and ab = 10, then (a + b)^2 = 9 + 40 = 49, so a + b = 7.
13. a^2 + b^2 in terms of (a + b) and ab
a^2 + b^2 = (a + b)^2 − 2ab
a^2 + b^2 = (a − b)^2 + 2ab
Proof: Start from (a + b)^2 = a^2 + 2ab + b^2 and subtract 2ab from both sides. ✔
Check: a = 4, b = 3
- a^2 + b^2 = 25
- (7)^2 − 2(12) = 49 − 24 = 25 ✔
Use: If a + b = 6 and ab = 8, then a^2 + b^2 = 36 − 16 = 20.
14. a^3 + b^3 = (a + b)^3 − 3ab(a + b)
Proof: (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
Notice 3a^2b + 3ab^2 = 3ab(a + b). So:
(a + b)^3 = a^3 + b^3 + 3ab(a + b)
Move the 3ab(a + b) across: a^3 + b^3 = (a + b)^3 − 3ab(a + b) ✔
Check: a = 2, b = 1
- Left: 8 + 1 = 9
- Right: 27 − 3(2)(3) = 27 − 18 = 9 ✔
15. a^3 − b^3 = (a − b)^3 + 3ab(a − b)
Proof: Same idea with (a − b)^3 = a^3 − b^3 − 3ab(a − b). Move the last term across. ✔
Check: a = 3, b = 1
- Left: 27 − 1 = 26
- Right: 8 + 3(3)(2) = 8 + 18 = 26 ✔
16. a^2 − b^2 = (a − b)(a + b), and the “difference of squares” chain
a^4 − b^4 = (a^2 − b^2)(a^2 + b^2) = (a − b)(a + b)(a^2 + b^2)
Proof: Treat a^2 as one number and b^2 as another. Apply a^2 − b^2 to them: (a^2)^2 − (b^2)^2 = (a^2 − b^2)(a^2 + b^2). Then break a^2 − b^2 again. ✔
Check: a = 2, b = 1
- Left: 16 − 1 = 15
- Right: (1)(3)(5) = 15 ✔
17. a^3 + b^3 + c^3 − 3abc = (a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca)
Proof: Multiply the right side.
- a × (a^2 + b^2 + c^2 − ab − bc − ca) = a^3 + ab^2 + ac^2 − a^2b − abc − a^2c
- b × (...) = a^2b + b^3 + bc^2 − ab^2 − b^2c − abc
- c × (...) = a^2c + b^2c + c^3 − abc − bc^2 − ac^2
Add all three. Everything cancels except a^3 + b^3 + c^3 − 3abc ✔
Check: a = 1, b = 2, c = 3
- Left: 1 + 8 + 27 − 18 = 18
- Right: (6)(1 + 4 + 9 − 2 − 6 − 3) = 6 × 3 = 18 ✔
Special case: If a + b + c = 0, then a^3 + b^3 + c^3 = 3abc.
18. Sum of consecutive pattern: (a + b)(a^2 − ab + b^2) and (a + b + c)^3 idea
(a + b + c)^3 is long, so we use a shortcut:
(a + b + c)^3 = a^3 + b^3 + c^3 + 3(a + b)(b + c)(c + a)
Check: a = 1, b = 1, c = 1
- Left: 27
- Right: 1 + 1 + 1 + 3(2)(2)(2) = 3 + 24 = 27 ✔
| Identity | Result |
|---|---|
| (a + b + c)^2 | a^2 + b^2 + c^2 + 2ab + 2bc + 2ca |
| (x + a)(x + b) | x^2 + (a + b)x + ab |
| (a + b)^2 + (a − b)^2 | 2(a^2 + b^2) |
| (a + b)^2 − (a − b)^2 | 4ab |
| a^2 + b^2 | (a + b)^2 − 2ab |
| a^3 + b^3 | (a + b)^3 − 3ab(a + b) |
| a^3 − b^3 | (a − b)^3 + 3ab(a − b) |
| a^4 − b^4 | (a − b)(a + b)(a^2 + b^2) |
| a^3 + b^3 + c^3 − 3abc | (a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca) |
Practice Questions
- Expand (x + 2)(x + 5)
- If a + b = 8 and ab = 15, find a^2 + b^2
- If a − b = 2 and ab = 24, find (a + b)^2
- If a + b = 5 and ab = 6, find a^3 + b^3
- Expand (x + y + 1)^2
- Find the value of 1^3 + 2^3 + (−3)^3 − 3(1)(2)(−3)
Answers:
- x^2 + 7x + 10
- 64 − 30 = 34
- 4 + 96 = 100
- 125 − 3(6)(5) = 125 − 90 = 35
- x^2 + y^2 + 1 + 2xy + 2y + 2x
- a + b + c = 0, so the expression equals 0
Practice Questions
- Find (x + 4)^2
- Find 99^2 using (100 − 1)^2
- Find 103 × 97 using a^2 − b^2
- Expand (2x + 1)^3
Answers: 1) x^2 + 8x + 16 2) 9801 3) 9991 4) 8x^3 + 12x^2 + 6x + 1
Tips for Students
- Don’t memorize blindly. If you forget a formula, just multiply the brackets.
- Watch the signs. Minus × minus = plus.
- Always verify with small numbers like a = 2, b = 1.
- Cube pattern: the coefficients 1, 3, 3, 1 come from Pascal’s triangle. For (a − b)^3 the signs alternate: + − + −.