# BASIC ALGEBRA FORMULAS - SPECIAL PRODUCTS, PROOFS, LINKS TO OTHER FORMULAS

Basic Algebra Formulas
if you have not already done so.
There we have listed out all the
Algebra Formulas that need to be
remembered.

Here we present the proofs of the first
six Formulas and provide Links for the
proofs of the last six Formulas and also
for Solved Examples and Exercise problems
on application of those Formulas.

## Proofs of Basic Algebra Formulas in Polynomials :

### Proof of Basic Algebra Formula 1 :

To Prove:
(a + b)2 = a2 + 2ab + b2

L.H.S. = (a + b)2 = (a + b)(a + b) = a(a + b) + b(a + b)
= a.a + a.b + b.a + b.b = a2 + (ab + ab) + b2
= a2 + 2ab + b2 = R.H.S. (Proved.)

### Proof of Basic Algebra Formula 2 :

To Prove:
(a - b)2 = a2 - 2ab + b2

Here we can directly prove as in Formula 1 above.
(I leave this as exercise for the reader.)

or, we can make use of Formula 1 to prove this.

Replacing b in Formula 1 by (-b), we get
{a + (-b)}2 = a2 + 2a(-b) + (-b)2
⇒ (a - b)2 = a2 - 2ab + b2 (Proved.)

### Proof of Basic Algebra Formula 3 :

To Prove:
(a + b)(a - b) = a2 - b2

L.H.S. = (a + b)(a - b) = a(a - b) + b(a - b)
= a.a - a.b + b.a - b.b
= a2 + (-ab + ab) - b2 = a2 - b2 = R.H.S. (Proved.)

### Proof of Basic Algebra Formula 4 :

Proving one of the Basic Algebra Formulas :

To Prove:
(a + b)(a2 - ab + b2) = a3 + b3

L.H.S. = (a + b)(a2 - ab + b2)
= a(a2 - ab + b2) + b(a2 - ab + b2)

= a(a2) - a(ab) + a(b2) + b(a2) - b(ab) + b(b2)
= a3 - a2b + ab2 + ba2 - b2a + b3

= a3 + (- a2b + ba2) + (ab2 - b2a) + b3
The terms in brackets are same and cancel each other.
∴ L.H.S. = a3 + b3 = R.H.S. (Proved.)

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### Proof of Basic Algebra Formula 5 :

To Prove:
(a - b)(a2 + ab + b2) = a3 - b3

Here we can directly prove as in Formula 4 above.
(I leave this as exercise for the reader.)
or, we can make use of Formula 4 to prove this.

Replacing b in Formula 4 by (-b), we get
{a + (-b)}{a2 - a(-b) + (-b)2} = a3 + (-b)3
⇒ (a - b)(a2 + ab + b2) = a3 - b3 = R.H.S. (Proved.)

### Proof of Basic Algebra Formula 6 :

To Prove:
(a + b)3 = a3 + 3a2b + 3ab2 + b3

L.H.S. = (a + b)3 = (a + b)(a + b)2
Using Formula 1 here, we get
L.H.S. = (a + b)(a2 + 2ab + b2)
= a(a2 + 2ab + b2) + b(a2 + 2ab + b2)
= a(a2) + a(2ab) + a(b2) + b(a2) + b(2ab) + b(b2)
= a3 + 2a2b + ab2 + ba2 + 2ab2 + b3
= a3 + (2a2b + ba2) + (ab2 + 2ab2) + b3
= a3 + 3a2b + 3ab2 + b3 = R.H.S. (Proved.)

## Proofs of Basic Algebra Formulas 7 to 12 :

Here is the Link for the proofs
of the remaining Formulas.

Proofs of 6 to 12 Formulas.

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## Solved Examples and Exercise problems : Basic Algebra Formulas

Here are the Links for the
Solved Examples and Exercise problems
on application of these Formulas.

Set 1 of Solved Examples and Exercise problems

Set 2 of Solved Examples and Exercise problems

Set 3 of Solved Examples and Exercise problems

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