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Algebra Formulas and Expression with Example

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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 = a4 + 4a3b + 6a2b2 + 4ab3 + b4)
  • (a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4)
  • a4 – b4 = (a – b)(a + b)(a2 + b2)
  • a5 – b5 = (a – b)(a4 + a3b + a2b2 + ab3 + b4)
  • If n is a natural number, an – bn = (a – b)(an-1 + an-2b+…+ bn-2a + bn-1)
  • If n is even (n = 2k), an + bn = (a + b)(an-1 – an-2b +…+ bn-2a – bn-1)
  • If n is odd (n = 2k + 1), an + bn = (a + b)(an-1 – an-2b +…- bn-2a + bn-1)
  • (a + b + c + …)2 = a2 + b2 + c2 + … + 2(ab + ac + bc + ….
  • Laws of Exponents
    (am)(an) = am+n
    (ab)m = ambm
    (am)n = amn
  • Fractional Exponents
    a0 = 1
    aman=am−naman=am−n
    amam = 1a−m1a−m
    a−ma−m = 1am
\[(a+b)^{2}=a^2+2ab+b^{2}\]
\[(a-b)^{2}=a^{2}-2ab+b^{2}\]
\[\left (a + b \right ) \left (a – b \right ) = a^{2} – b^{2}\]
\[\left (x + a \right )\left (x + b \right ) = x^{2} + \left (a + b \right )x + ab\]
\[\left (x + a \right )\left (x – b \right ) = x^{2} + \left (a – b \right )x – ab\]
\[\left (x – a \right )\left (x + b \right ) = x^{2} + \left (b – a \right )x – ab\]
\[\left (x – a \right )\left (x – b \right ) = x^{2} – \left (a + b \right )x + ab\]
\[\left (a + b \right )^{3} = a^{3} + b^{3} + 3ab\left (a + b \right )\]
\[\left (a – b \right )^{3} = a^{3} – b^{3} – 3ab\left (a – b \right )\]
\[ (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2xz\]
\[ (x + y – z)^{2} = x^{2} + y^{2} + z^{2} + 2xy – 2yz – 2xz\]
\[ (x – y + z)^{2} = x^{2} + y^{2} + z^{2} – 2xy – 2yz + 2xz\]
\[ (x – y – z)^{2} = x^{2} + y^{2} + z^{2} – 2xy + 2yz – 2xz\]
\[ x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz -xz\]
\[ x^{2} + y^{2} = \frac{1}{2} \left [(x + y)^{2} + (x – y)^{2} \right ]\]
\[ (x + a) (x + b) (x + c) = x^{3} + (a + b +c)x^{2} + (ab + bc + ca)x + abc\]
\[ x^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2})\]
\[ x^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2})\]
\[ x^{2} + y^{2} + z^{2} -xy – yz – zx = \frac{1}{2} [(x-y)^{2} + (y-z)^{2} + (z-x)^{2}]\]
\[\mathbf{a_{1}x + b_{1}y + c_{1} = 0}\]
\[\mathbf{a_{2}x+ b_{2}y + c_{2}} = 0 \]
Distributive Propertya ( b+c) = (a × b) + (a × c)
Commutative Property of Additiona + b = b+a
Commutative Property of Multiplicationa× b = b×a
Associative Property of Additiona + (b + c ) = ( a+ b ) +c
Associative Property of Multiplicationa ( b ×  c ) = ( a× b ) × c
Additive Identity Propertya +0 = a
Multiplicative Identity Propertya×1 = a
Additive Inverse Propertya+ ( -a) = 0
Multiplicative Inverse Propertya × 1/a = 1
Zero Property of Multiplicationa × 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=”latex-for-amp”>\[ \vec{0}\] is the <a href=”https://byjus.com/maths/additive-identity-vs-multiplicative-identity/”>additive identity</a> in vector addition.
\[\vec{a}+\left ( -\vec{a} \right )=-\vec{a}+\vec{a}=\vec{0}\], therefore <span class=”latex-for-amp”>\[\vec{a}\]  is the inverse in vector addition.

What is Algebra?

Algebra is a mathematical segment that replaces letters for the number. An algebraic equation is working as a scale where both sides represent the same things and numbers are taken as the constants. Algebra may include multiple mathematical terms like the real number, complex numbers, vectors, matrix, and many other forms.

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.

Algebra Formulas

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 x-38 = 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 52 – 32

Solution:
Using the formula a2 – b2 = (a – b)(a + b)
where a = 5 and b = 3
(a – b)(a + b)
= (5 – 3)(5 + 3)
= 2 ×× 8
= 16

Question 2: 
43 ×× 42 = ?

Solution:
Using the exponential formula (am)(an) = am+n
where a = 4
43 ×× 42
= 43+2
= 45
= 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 problem-solving 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 tech-related 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.

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