Math
Algebra Formulas and Expression with Example
Table of Contents
Table of Content
 Algebra Formulas
 What is Algebra?
 Algebra Equations
 Algebra Problems with a solution
 Why Algebra Formula Needs etc.?
Algebra Formulas
A basic formula in Algebra represents the relationship between different variables. The variable could be taken as x, y, a, b, c or any other alphabet that represents a number unknown yet. Example – (x + y = z)
 (a + b)^{4} = a^{4} + 4a^{3}b + 6a^{2}b^{2} + 4ab^{3} + b^{4})
 (a – b)^{4} = a^{4} – 4a^{3}b + 6a^{2}b^{2} – 4ab^{3} + b^{4})
 a^{4} – b^{4} = (a – b)(a + b)(a^{2} + b^{2})
 a^{5} – b^{5} = (a – b)(a^{4} + a^{3}b + a^{2}b^{2} + ab^{3} + b^{4})
 If n is a natural number, a^{n} – b^{n} = (a – b)(a^{n1} + a^{n2}b+…+ b^{n2}a + b^{n1})
 If n is even (n = 2k), a^{n} + b^{n} = (a + b)(a^{n1} – a^{n2}b +…+ b^{n2}a – b^{n1})
 If n is odd (n = 2k + 1), a^{n} + b^{n} = (a + b)(a^{n1} – a^{n2}b +… b^{n2}a + b^{n1})
 (a + b + c + …)^{2} = a^{2} + b^{2} + c^{2} + … + 2(ab + ac + bc + ….
 Laws of Exponents
(a^{m})(a^{n}) = a^{m+n}(ab)^{m} = a^{m}b^{m}(a^{m})^{n} = a^{mn}  Fractional Exponents
a^{0} = 1
aman=am−naman=am−n
amam = 1a−m1a−m^{}a−ma−m = 1am
\[(a+b)^{2}=a^2+2ab+b^{2}\] 
\[(ab)^{2}=a^{2}2ab+b^{2}\] 
\[\left (a + b \right ) \left (a – b \right ) = a^{2} – b^{2}\] 
\[\left (x + a \right )\left (x + b \right ) = x^{2} + \left (a + b \right )x + ab\] 
\[\left (x + a \right )\left (x – b \right ) = x^{2} + \left (a – b \right )x – ab\] 
\[\left (x – a \right )\left (x + b \right ) = x^{2} + \left (b – a \right )x – ab\] 
\[\left (x – a \right )\left (x – b \right ) = x^{2} – \left (a + b \right )x + ab\] 
\[\left (a + b \right )^{3} = a^{3} + b^{3} + 3ab\left (a + b \right )\] 
\[\left (a – b \right )^{3} = a^{3} – b^{3} – 3ab\left (a – b \right )\] 
\[ (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2xz\] 
\[ (x + y – z)^{2} = x^{2} + y^{2} + z^{2} + 2xy – 2yz – 2xz\] 
\[ (x – y + z)^{2} = x^{2} + y^{2} + z^{2} – 2xy – 2yz + 2xz\] 
\[ (x – y – z)^{2} = x^{2} + y^{2} + z^{2} – 2xy + 2yz – 2xz\] 
\[ x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz xz\] 
\[ x^{2} + y^{2} = \frac{1}{2} \left [(x + y)^{2} + (x – y)^{2} \right ]\] 
\[ (x + a) (x + b) (x + c) = x^{3} + (a + b +c)x^{2} + (ab + bc + ca)x + abc\] 
\[ x^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2})\] 
\[ x^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2})\] 
\[ x^{2} + y^{2} + z^{2} xy – yz – zx = \frac{1}{2} [(xy)^{2} + (yz)^{2} + (zx)^{2}]\] 
\[\mathbf{a_{1}x + b_{1}y + c_{1} = 0}\] 
\[\mathbf{a_{2}x+ b_{2}y + c_{2}} = 0 \] 
Distributive Property  a ( b+c) = (a × b) + (a × c) 
Commutative Property of Addition  a + b = b+a 
Commutative Property of Multiplication  a× b = b×a 
Associative Property of Addition  a + (b + c ) = ( a+ b ) +c 
Associative Property of Multiplication  a ( b × c ) = ( a× b ) × c 
Additive Identity Property  a +0 = a 
Multiplicative Identity Property  a×1 = a 
Additive Inverse Property  a+ ( a) = 0 
Multiplicative Inverse Property  a × 1/a = 1 
Zero Property of Multiplication  a × 0 = 0 
If\[\vec{a}=x\hat{i}+y\hat{j}+z\hat{k}\] then magnitude or length or norm or absolute value of \[\vec{a} \] is \[ \left  \overrightarrow{a} \right =a=\sqrt{x^{2}+y^{2}+z^{2}}\] 
A vector of unit magnitude is unit vector. If \[\vec{a}\] is a vector then unit vector of \[\vec{a}\] is denoted by \[\hat{a}\] and \[\hat{a}=\frac{\hat{a}}{\left  \hat{a} \right }\] Therefore \[ \hat{a}=\frac{\hat{a}}{\left  \hat{a} \right }\hat{a}\] 
Important unit vectors are \[\hat{i}, \hat{j}, \hat{k}\], where \[\hat{i} = [1,0,0],\: \hat{j} = [0,1,0],\: \hat{k} = [0,0,1]\] 
If \[ l=\cos \alpha, m=\cos \beta, n=\cos\gamma,\] then \[ \alpha, \beta, \gamma,\] are called directional angles of the vectors\[\overrightarrow{a}\] and \[\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1\] 
\[\vec{a}+\vec{b}=\vec{b}+\vec{a}\] 
\[\vec{a}+\left ( \vec{b}+ \vec{c} \right )=\left ( \vec{a}+ \vec{b} \right )+\vec{c}\] 
\[k\left ( \vec{a}+\vec{b} \right )=k\vec{a}+k\vec{b}\] 
\[\vec{a}+\vec{0}=\vec{0}+\vec{a}\], therefore <span class=”latexforamp”>\[ \vec{0}\] is the <a href=”https://byjus.com/maths/additiveidentityvsmultiplicativeidentity/”>additive identity</a> in vector addition. 
\[\vec{a}+\left ( \vec{a} \right )=\vec{a}+\vec{a}=\vec{0}\], therefore <span class=”latexforamp”>\[\vec{a}\] is the inverse in vector addition. 
What is Algebra?
Further, algebra is broken down into parts – one is elementary algebra and the other is abstract algebra. The first one is suitable for medical, science, engineering, mathematics, or economics, etc. At the same time, the second type is suitable for advanced mathematics only.
Basic Algebra Equations
A basic equation in Algebra is an expression that contains at least one variable that we need to calculate to find something unknown to you. Example – (x – 38 = 12). This is an algebraic equation where we need to calculate the unknown variable x.
Algebra Problems with a Solution
Algebra is used everywhere and it can be used for solving daily problems too when it comes to knowing about the unknown variable. Let us see how it works actually. Take an example where you are trying to calculate the balls you started the day with if you sold 38 but still 12 is remaining.
Here, the algebraic equation would be x38 = 12. We are trying to calculate unknown variable x that will tell you how many balls you started your day. The solution to the problem is (x=50), it means you started your day with 50 balls.
Question 1: Find out the value of 5^{2} – 3^{2}
Solution:
Using the formula a^{2} – b^{2} = (a – b)(a + b)
where a = 5 and b = 3
(a – b)(a + b)
= (5 – 3)(5 + 3)
= 2 ×× 8
= 16
Question 2: 4^{3} ×× 4^{2} = ?
Solution:
Using the exponential formula (a^{m})(a^{n}) = a^{m+n}where a = 4
4^{3} ×× 4^{2}= 4^{3+2}= 4^{5}= 1024
Why Algebra Formula Needs?
Wondered how Algebra formulas are needed even outside your high school? In reality, they are used everywhere either it is related to bill payments, calculation of healthcare costs, your future investment, planning for a house on EMIs or budget management for your project, etc. A deep understanding of Algebra formulas or equations can develop critical thinking, makes your logics stronger, and enhances your problemsolving skills especially when you enter the workspace.
Ultimately, the person having a good understanding of Algebra had a greater opportunity of getting success in science, mathematics, engineering, economics, complex mathematics fields or any other techrelated fields, etc. We wish you a successful future ahead with the right skills and knowledge of various Algebraic equations or formulas.
You should remember all the algebra Formulas which helps you to solve algebra equations and easy to understand. You should also check the more Maths formulas which are the help to learn more. if you have any questions regarding algebra related and algebra equations, you should share your problem here for the solution.
List of Maths Formulas By Class

Math2 years ago
Surface Area of a Triangular Prism Formula & Volume of a Triangular

Math2 months ago
Average Rate Of Change Formula Made Simple

Math2 months ago
Math Formulas Made Simple: A StepbyStep Guide

Math2 years ago
Percentage Formulas  How to Calculate Percentages of a Number?

Math1 year ago
Triangular Pyramid Formula  Volume & Surface Area of a Triangular Pyramid

Math2 years ago
What is Integration? List of Integration by Parts Formulas

Math2 years ago
List of Pyramid Formula – Surface Area, Volume of Pyramid

Math2 years ago
Surface Area of a Rectangular Prism Formula & Volume of a Rectangular