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& testbook.com, , Algebraic Identities: Know All The Identities Of Algebra With, Their Application, , 2021/04/21, , Algebraic Identities are the fundamentals of Algebra. These identities are the basics that everyone must, know as it is much easier to solve algebraic linear equations using identities. Algebraic identity is a must, for candidates who are appearing for competitive or entrance exams because it saves a lot of time if you, solve questions using these identities. In this article, we will cover all the types of linear identities along, with the type of questions where they are applicable. Also, we will list some tips that are helpful for, solving questions related to algebraic identities. Read the article below to get a better understanding of, algebraic identities with the help of solved sample questions., , What are Algebraic Identities?, , An algebraic identity is an algebraic equation that always remains true for any value of variables in it. Or, one can say that an algebraic identity is simply equality that does not change with the change in the, values of the variables. Algebraic identities are often used to factorize polynomials in a way that is easier, and faster. In this article, we will list all these identities along with their application in various types of, questions., , History of Algebraic Identities, , Algebra was developed in several stages and the ancient Babylonians were the ones to develop the first, stage, ie, Rhetorical Algebra. Later, Diophantus and the Indian Mathematician Brahmagupta invented, geometric algebra. However, they were unable to reach the static equation-solving stage. I was AlKhwarizmi, a medieval Persian mathematician who was able to reach this stage where the objective was, to find those numbers that satisfy a particular relation (equation). This is how later, Algebraic identity, came into being., , If you’ve learned Algebraic Identities, you can move on to learn about Linear Equations in one variable, concept here!, , Standard Algebraic Identities List, , The basic theorem of algebra says that the range of the complex numbers is closed algebraically, that is,, , all polynomial equations with complex coefficients plus degrees at least one hold a solution.

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Some of the standard identities applied and practised in various branches of mathematics are listed, below. All these identities are derived by employing the Binomial theorem. Will discuss the theorem in, , the coming headings., , Standard Identity-1: Algebraic identity of the square of the summation of two terms is:, , (a+b)? = a? + b? + 2ab, , Standard Identity-2: Algebraic identity of the square of the difference of two given terms is:, , (a — b)? = a2 + Bb? — 2ab, , Standard Identity-3: Algebraic identity of the difference of two squares is:, a? — b? = (a+ b) (a —b), , Standard Identity-4: Algebraic identity for (x+a)(x+b) is:, (x + a)(a + b) = x + x(a + b) + ab, , Different Types of Algebraic Identities, Different types of algebraic identities are listed below:, , Factoring Identities, , The following identities are useful when you need to perform factorization in an algebraic equation:, a2 — b= (a + b)(a— bd), , a3 — b3 = (a — b)(a? + ab + 6?), , a3 + B38 = (a + b)(a? — ab + 6?), , at — b4 = (a2 — b?)(a? + B?), , (x + a)(a +b) = 2? + (a+ b)a + ab, , oe © NS oP, , Three Variable Identities, , Following are the algebraic identities that use 3 variables in a single equation:, , (a+b+c)? =a? +b? +c? + 2ab + 2be + 2ca, a3 + B38 + c8—3abe = (a + b + c)(a? + b? + c?—ab-be-ca), , Proof of Algebraic Identities, , So far we read about the various algebraic identities along with the history and various types. Let us now, , learn how to prove these identities taking some of them as examples., , «= « gfa@bkb2— e241. 95h 2

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£LO0LOLb® — * re, , The formula of (a+b)? is identical to (a + b) x (a + b). This can be seen as a square whose sides are (a + b), , and area is (@ + 8)”, , & testbook, , , , 2, The area of the square(% a b) in terms of the product is calculated as (a+b)(a+b). Apart from this the, 2, area of the square (a +b) is also identical to the summation of the areas of the individual squares and, , 2 = @ 2, rectangles. Therefore, we can write(@ + 5)? = a? + 2ab + 6?, , Proof of (4 ~ 6)? = a? — 2ab + b?, , (asp, , Let us consider as the area of a square with measure (a —b)., , 2, To surmise this, let’s start with a bigger square of area “. We will decrease the length of each side by b., 2 —b)2, We now have to eliminate the extra bits from ~ to be left with (4 ) ‘, , —p)2, Consider the below figure, (a— 6) and look at the larger green area., , a-b b, , , , , |, , To receive the blue square from the larger green square, we must subtract the vertical and horizontal, , © — $$, a, , sections that have the area ab. Nevertheless, removing ab twice will also eliminate the overlapping, , b, , 2, square at the bottom right edge twice. Therefore, we join ~ . Hence the formula becomes,, , (a — b)? = a? — 2ab +b?, , Proof of (* + @)(@ +b) =a? + (a +b) + ab

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(x + a)(x + b) can be represented as the area of a rectangle whose sides are x + aand x + b., , ® testbook, xt+b, , St, , , , x b, , The area of a rectangle with sides x + a and x + b in terms of the individual areas of the rectangles and, the square is a? + aa + bx + ab = a? + (a+ b)a + ab, , (a + a)(a +6) = 2? + x(a +b) + ab,, , . Therefore,, , Proof of 2-0? = (a + 6)(a-b), , , , Consider the above square figure whose side is a=(a—b)+b units. In this case, the bigger square is divided, into three quadrilaterals, a combination of rectangles and squares., , ide)2, Applying the area of a rectangle and area of the square formula i.e. (lengthxbreadth) and (side), , respectively we can picturise the identity as shown. The total area of the bigger square is the summation, of areas of rectanglest+squares involved in it., , a? = a(a—b)+b(a—b) +b, a? = (a+b)(a— 6b) + B2, a2—b? = (a + b)(a-b), , Proof of (@+ 6+ ¢)? =a? +6? +c? + 2ab + 2be + 2ca

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To prove the above formula, take a square with side a+b+c units, as shown in the figure. The bigger, , square is separated into nine quadrilaterals i.e. a combination of rectangles+squares., , ©, ° °, + +, a 2, + Ole, © ©, °, , , , at+btc, , ALND, Applying the area of a rectangle and area of the square formula i.e. (lengthxbreadth) and (side), respectively we can picturise the identity as shown. The total area of the bigger square is the summation, of areas of rectanglest+squares involved in it., , (a+b+c)?=a?+ab+ac+ab+b?+ be +ac+ be + c?, , (a+b+c)? = a? +b? + c+ 2ab + 2be + 2ca, , Similarly we can prove other formulas as well., , Methods to Verify Algebraic Identities, , In the previous heading, we learnt how we can prove the various algebraic identities. Let us now proceed, , towards the methods to verify algebraic identities., Applying Substitution Method, , As the name implies, substitution naturally means placing numbers or weights in the place of variables, of the given equation. In this method, an arithmetic operation is executed by replacing the values for the, variables., , For example, if we have x-4=6, Consider the LHS first:, If we substitute x= 10, On the LHS then:, , x-4=10-4=6