March 26, 2023

# Algebra Formulas

Numbers and letters are both used in algebra. The algebraic formula’s unknown quantities are represented by letters or alphabets. Now, an equation or formula is created using a mix of integers, letters, factorials, fractions, decimals, log  etc. There are a few highly crucial algebraic formulae and equations that students must memorise while they prepare for their exams. The foundation of basic or elementary algebra is these formulas. Knowing the formulas alone is insufficient. you need to know how to use these formulas to solve problems.
Here, we’ll list all the significant algebraic formulas. The extensive list will make it possible for the students to look over it quickly before tests or to refer to it at any time they like.

## Important Algebra Formulas  (List of all Algebraic Identities)

An algebraic identity is an algebraic equation which is true for all values of the variable (s).

• $(a+b)^{2}=a^{2}&space;+b^{2}+2ab$
• $(a-b)^{2}=a^{2}&space;+b^{2}-2ab$
• $(a+b)^{2}=(a-b)^{2}+4ab$
• $(a-b)^{2}=(a+b)^{2}-4ab$
• $a^{2}+b^{2}=&space;\frac{1}{2}[(a+b)^{2}+(a-b)^{2}]$
• $a^{2}-b^{2}=&space;(a-b)(a+b)$
• $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$
• $(a-b)^{3}=a^{3}-b^{3}-3ab(a-b)$
• $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$
• $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$
• $(a+b+c)^{2}=a^{2}&space;+b^{2}+c^{2}+2ab+2bc+2ca$
• $(a-b-c)^{2}=a^{2}&space;+b^{2}+c^{2}-2ab-2bc-2ca$
• $(a+b+c)^{3}=a^{3}&space;+b^{3}+c^{3}+3(a+b)(b+c)(c+a)$
• $a^{3}&space;+b^{3}+c^{3}-3abc=&space;(a+b+c)(a^{2}&space;+b^{2}+c^{2}-ab-bc-ca)$
• If $a+b+c=0&space;\,&space;\,&space;\mathrm{then&space;}\,&space;\,&space;a^{3}&space;+b^{3}+c^{3}=3abc$
• $a^{3}+b^{3}+c^{3}-3abc=\frac{1}{2}(a+b+c)[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]$
• $\mathrm{If&space;\,&space;\,&space;a^{3}+b^{3}+c^{3}=3abc&space;\,&space;then&space;\,&space;\,&space;either&space;\,&space;\,&space;a+b+c=o&space;\,&space;\,&space;or\,&space;\,&space;a=b=c&space;}$
• $\mathrm{If&space;\,&space;\,&space;a^{3}+b^{3}+c^{3}=ab+bc+ca&space;\,&space;\,&space;then\,&space;\,&space;a=b=c}$
• $a^{4}+b^{4}+a^{2}b^{2}=(a^{2}+b^{2}+ab)(a^{2}+b^{2}-ab)$
• $a^{4}-b^{4}=(a-b)(a+b)(a^{2}+b^{2})$
• $a^{8}-b^{8}=(a-b)(a+b)(a^{2}+b^{2})(a^{4}+b^{4})$
• $If&space;\,&space;\,&space;a+\frac{1}{a}=&space;x&space;\,&space;\,&space;then&space;\,&space;\,&space;a^{3}+\frac{1}{a^{3}}&space;=x^{2}-3x$
• $If&space;\,&space;\,&space;a-\frac{1}{a}=&space;x&space;\,&space;\,&space;then&space;\,&space;\,&space;a^{3}-\frac{1}{a^{3}}&space;=x^{2}+3x$
•  If “$n$” is a natural number then  $a^{n}&space;-&space;b^{n}&space;=&space;(a-b)&space;(a^{(n-1)}&space;+&space;a^{(n-2)}b&space;+....+b^{(n-2)}a&space;+&space;b^{n-1})$
• If “$n$” is a even number  then   $a^{n}&space;+&space;b^{n}&space;=&space;(a+b)&space;(a^{(n-1)}&space;-&space;a^{(n-2)}b&space;+.....+&space;b^{(n-2)}a&space;-&space;b^{(n-1))})$
• If “$n$” is an odd number  then  $a^{n}&space;+&space;b^{n}&space;=&space;(a-b)&space;(a^{(n-1)}&space;-a^{(n-2)}b&space;+....&space;-&space;b^{(n-2)}a&space;+&space;b^{(n-1)})$
• $If&space;\,&space;\,&space;\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+........\infty&space;}}}}&space;\,&space;\,&space;where&space;\,&space;\,&space;x=&space;n(n+1)$  then  $If&space;\,&space;\,&space;\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+........\infty&space;}}}}&space;=n+1$
• $If&space;\,&space;\,&space;\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-........\infty&space;}}}}&space;\,&space;\,&space;where&space;\,&space;\,&space;x=&space;n(n+1)$ then $If&space;\,&space;\,&space;\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-........\infty&space;}}}}&space;=n$

### Remainder Theorem:

Let $p(x)=a_{0}+a_{1}x+a_{2}x^{2}+..........+a_{n}x^{n}$

be a polynomial of degree  $n\geq&space;1$, and let $a$ be any real number. When  is $p(x)$ divided by $(x-a)$, then the remainder is $p(a)$.

#### Factor theorem :

Let  $p(x)$ be a polynomial of degree greater than or equal to 1 and a be a real number such that $p(a)$ = 0 , then  $(x-a)$ is a factor of $p(x)$.

Conversely, if $(x-a)$ is a factor of  $p(x)$ , then $p(a)$ = 0 .

##### Componendo  and Dividendo rule :
•  $If&space;\,&space;\,&space;\frac{a}{b}=\frac{b}{c}&space;\,&space;\,&space;then&space;\,&space;\,&space;\frac{a+b}{a-b}=\frac{c+d}{c-d}$

• $If&space;\,&space;\,&space;\frac{a+b}{a-b}=\frac{c}{d}&space;\,&space;\,&space;then&space;\,&space;\,&space;\frac{a}{b}=\frac{c+d}{c-d}$
###### Binomial theorem:

$(a+b)^{n}=&space;^{n}C_{0}a^{n}b^{0}+&space;^{n}C_{1}a^{n-1}b^{1}+^{n}C_{2}a^{n-2}b^{2}+&space;.....+^{n}C_{n-1}a^{1}b^{n-1}+^{n}C_{n}a^{0}b^{n}$

where n is a positive number and $^{n}C_{r}=&space;\frac{n!}{r!(n-r)!}$  .

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