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Maths
Algebraic Expressions, Identities and Factorisation

Algebraic Expressions, Identities and Factorisation

Basic terms related to Algebraic Expressions, Identities and Factorisation

1.0Algebraic expression

A combination of constants and variables connected by the signs of fundamental operations of addition, subtraction, multiplication and division is called an 'Algebraic expression'. E.g., 3x,2x−3y etc., are algebraic expressions, the numbers 3,2,−3 are constants and x,y are variables.

2.0Important Definitions

  1. Variable: A quantity which can take different numerical values is called a variable. E.g., x,y,a,z, etc.
  2. Constant: A quantity having a fixed numerical value is called a constant.E.g., 3,5,8,–11 etc.
  3. Terms: Various parts of an algebraic expression which are separated by the signs + or - are called the 'terms' of the expression. E.g., (i) 3x2−7x+2, is an algebraic expression consisting three terms i.e., 3x2,−7x and 2 (ii) 2x3−3x2+4x−7 is an algebraic expression consisting four terms, i.e., 2x3,−3x2,4x,−7 E.g., We know that the perimeter 'P' of a square of side 's' is given by P=4×s. Here, ' 4 ' is a constant and ' P ' and 's' are variables.
  4. Factor: Terms themselves can be formed as the product of factors. E.g., (i) The term 4 x is the product of its factors 4 and x . (ii) The term 5 is made up of just one factor i.e., 5 .
  5. Constant term: The term of an algebraic expression having no literal factor is called its constant term. E.g., (i) In the expression 3a2−9, the constant term =−9 (ii) In the expression 5x2−6xy+7y−9y2+23​, the constant term =23​
  6. Coefficient: Any factor of a term of an algebraic expression is called the coefficient of the remaining factor of the term. E.g., (i) In −5xy, the coefficient of ' x ' is −5y; the coefficient of ' y ' is −5x and the coefficient of ' xy ' is -5 . (ii) In −x, the coefficient of ' x ' is -1 . (iii) In 3a2bc, the coefficient of ' a2 ' is 3bc, the coefficient of ' b ' is 3a2c and the coefficient of ' c ' is 3a2b.

Numerical coefficient

The numerical factor of a term of an algebraic expression is called numerical coefficient. E.g., In −10x2y, numerical coefficient is -10 .

-If there is no numerical coefficient then it is considered as 1 . E.g., In x2y the coefficient of x2y=1

  • Q. Identify the terms, their numerical coefficients for each of following expressions : (i) 4x3y2−4xy+1 (ii) 0.3a−0.6ab+9 Explanation :
ExpressionTermsNumerical coefficient
(i)4x3y2−4xy+14x3y24
−4xy-4
1
(ii)0.3a−0.6ab+90.3 a0.3
-0.6 ab-0.6
9

3.0Like And Unlike Terms

Like Terms: The terms having the same literal factors with the same exponents are called like or similar terms. E.g., 2x2y,x2y,−3x2y are like terms as they contain two variables x and y with equal exponent.

Unlike Terms: Terms having different literal factors are called unlike terms. E.g., 2x2y and 2xy2 are unlike terms as the factors of 2x2y are 2,x,x and y, whereas factors of 2xy2 are 2,x,y and y which are not same. like terms E.g.,  unlike terms 4xy,5x2,−4x2,9x3​

  • Non-negative integers are whole numbers i.e., 0,1,2,3,4,
  • An algebraic expression in x is said to be in standard form when the terms are written either in increasing order or decreasing order of the powers of x in various terms.

4.0Polynomials

An algebraic expression in which the variables involved have only non-negative integral power, is called a polynomial. An algebraic expression of the form p(x)=an​xn+an−1​xn−1+an−2​xn−2+…….....+a11​x1+a0​x0 where an​=0 and a0​,a1​,a2​,…...an​ are real numbers and each power of x is a non- negative integer is called a polynomial. Hence, an​,an−1​,an−2​….. a a0​ are coefficient of xn,xn−1,xn−2. x0. and an​xn,an−1​Xn−1,an−2​Xn−2, are terms of the polynomial. E.g., (i) p(x)=an​Xn+an−1​Xn−1+…..a1​x+a0​ is a polynomial in variable x . (ii) p(y)=an​yn+an−1​yn−1+…..a1​y+a0​ is a polynomial in variable y. (iii) p(z)=an​zn+an−1​zn−1+…..a1​Z+a0​ is a polynomial in variable z. All expressions which are written above in E.g. (i), (ii) and (iii) are polynomials. E.g., (i) p(u)=21​u3−3u2+2u−4 is a polynomial in variable u. Here, 21​u3,−3u2,2u,−4 are the terms of the above polynomial and 21​,−3,2,−4 are constants.

5.0Degree of Polynomial

Degree of the polynomial in one variable is the highest exponent of that variable in the expression. E.g., 3x4+5x3+7x2+8

Degree =4 (highest exponent) For polynomials in two or more variables, the degree of a term is sum of the exponents of the variables in that term and degree of the polynomial is the highest of degrees of all the terms in the polynomial. E.g., 5x3yz+6x4+23

Degree =3+1+1=5 (highest exponent of a term)

0 is called a zero polynomial and its degree is not defined.

6.0Classification of Polynomials

Based on number of terms

  1. Monomial : An algebraic expression containing only one term is called a monomial. E.g., 7,3y,5xy,−5,7xy,32​x2yz,35​a2bc3 etc. are all monomials.
  2. Binomial : An algebraic expression containing two unlike terms is called a binomial. E.g., 3x+2y,7xyz−5,2x−3,xyz−5 etc. are all binomials.
  3. Trinomial : An algebraic expression containing three unlike terms is called a trinomial. E.g., x3−2y3+3x2y2z,7+a+5b etc. are all trinomials.
  4. Multinomial : An expression is said to be a multinomial if it contains more than three terms.
  • For Binomial or trinomial, the terms must be unlike as like terms will get added or subtracted. So, the number of terms will change. E.g., 2x+4x is not a binomial it is a monomial as 2x+4x=6x

Based on Degree

  1. Linear Polynomial : A polynomial of degree 1 is called as linear polynomial. General form : ax+b,(a=0) E.g., y+3,21​+a etc. are linear polynomial.
  2. Quadratic Polynomial : A polynomial of degree 2 is called as quadratic polynomial. General form : a2+bx+c,(a=0) E.g., 2y2+5y−2,5−23​x2+3y etc.
  3. Cubic Polynomial : A polynomial of degree 3 is called as cubic polynomial. General form : ax3+bx2+cx+d,(a=0) E.g., 7x3−3x2+8x+10,67y3 etc. are cubic polynomial.
  4. Biquadratic Polynomial : A polynomial of degree 4 is called as biquadratic polynomial. General form : ax4+bx3+cx2+dx+e,(a=0) E.g., 5x4−23x3+18x2+11x−5.
  5. Constant Polynomial : A polynomial of degree 0 is called constant polynomial. E.g., 3,−7,21​,5​ etc. are constant polynomial.
  • Q. Which type of polynomial is x2y2+2xy+1−xy ? Explanation : The given polynomial can be written as x2y2+2xy−xy+1=x2y2+xy+1 This is a trinomial, as it has 3 unlike terms. Polynomial of degree 4 is called as quadratic polynomial and of degree 5 is called as quintic polynomial.

7.0Addition Of Algebraic Expressions

Like terms can be added but unlike terms cannot be added. For addition of two or more algebraic expressions, we first collect like terms and then find the sum of coefficients of these terms.

  • Q. Add : 6x2−5x+6,−5x2+6x+1 and 9x2−x+8 Explanation : 6x2+(−5x2)+9x2−5x+6x−x+6+1+8 =(6−5+9)x2+(−5+6−1)x+15 [Collecting like terms] =10x2+0⋅x+15=10x2+15 [Adding like terms]
  • Q. Add : 5x2−31​x+25​,−21​x2+21​x−31​ and −2x2+51​x−61​ Solution: Required sum =(5x2−31​x+25​)+(−21​x2+21​x−31​)+(−2x2+51​x−61​) =5x2−21​x2−2x2−31​x+21​x+51​x+25​−31​−61​ [Collecting like terms] =(5−21​−2)x2+(−31​+21​+51​)x+(25​−31​−61​) [Adding like terms] =(210−1−4​)x2+(30−10+15+6​)x+(615−2−1​)=25​x2+3011​x+2

8.0Subtraction Of Algebraic Expressions

Like terms can be subtracted but unlike terms cannot be subtracted. For subtraction of two or more algebraic expressions, we first collect like terms and then find the difference of coefficients of these terms.

  • Q. Subtract −9x2+6x−8 from 3x2−4x−5. Explanation : 3x2−4x−5−(−9x2+6x−8) =3x2−4x−5+9x2−6x+8 =3x2+9x2−4x−6x−5+8 [Collecting like terms] =(3+9)x2+(−4−6)x+3 [Adding like terms] =12x2+(−10)x+3 =12x2−10x+3

9.0Multiplication Of Algebraic Expressions

To multiply two or more monomials we use the laws of exponents and the rules of signs with the commutative and associative properties of multiplication.

MultiplicationResultant Sign
(+)×(+)+
(−)×(−)+
(+)×(−)-
(−)×(+)-
  • am×an=am+n
  • (am)n=amn
  • am÷an=am−n

While multiplying algebraic expression we will use these above mentioned concepts.

  • We use the rules of exponents and powers when we multiply variables.

Multiplication of a monomial by a monomial

Rule: Product of two or more monomials = (Product of their numerical coefficients along with their signs) × (Product of their variable parts)

Multiplying two monomials

  • Q. Find the product of 20x10y20z30 and (10xyz)2. Explanation : We have, 20x10y20z30,(10xyz)2 =(20x10y20z30)×(10xyz)×(10xyz) =(20×10×10)×(x10×x×x×y20×y×y×z30×z×z) =2000x10+1+1y20+1+1z30+1+1 =2000x12y22z32
  • Q. Find the product of (i) 6xy and −3x2y2 (ii) (4a​)2×(−41​a2) (iii) 4x2yz and −23​x2yz2 (iv) 2.1a2bc and 4ab2 Solution : (i) 6xy and −3x2y2 ={6×(−3)}×{xy×x2y2}=−18×{x×x2×y×y2}=−18x3y3 (ii) (4a​)2×(−41​a2) =42a2​×(−41​a2) =161​a2×(−41​a2) =161​×(−41​)×a2×a2 =16×4−1​×a4=64−1​a4 (iii) 4x2yz and −23​x2yz2 We have, (4x2yz)×(−23​x2yz2)=(4×−23​)×(x2×x2×y×y×z×z2) =−6x2+2y1+1z1+2=−6x4y2z3 (iv) 2.1a2bc and 4ab2 We have, (2.1a2bc)×(4ab2) =(2.1×4)×(a2×a×b×b2×c) =8.4a2+1 b1+2c=8.4a3 b3c

Multiplying three or more monomials

  • Q. Find the volume of the rectangular boxes with following length, breadth and height : Length = 2ax, Breadth =3 by, Height =5cz Explanation : We know that the volume of a rectangular box = Length × Breadth × Height Volume =2ax×3by×5cz =(2×3×5)×(ax×by×cz)=30 abcxyz

Multiplying a monomial by a polynomial

Rule: Multiply each term of the binomial by the monomial, using the distributive law, a×(b+c)=a×b+a×c ' a ' is monomial and (b+c ) is binomial.

  • Q. Multiply : 3x and (4x+3y2) Explanation : =3x×(4x+3y2) =3x×4x+3x×3y2 =12x2+9xy2
  • Q. Find the product of (43​x2yz2,0.5xy2z2,1.16x2yz3,2xyz) Solution : We have,
(43​x2yz2)×(0.5xy2z2)×(1.16x2yzz3)×(2xyz)=(43​×0.5×1.16×2)×(x2×x×x2×x×y×y2×y×y×z2×z2×z3×z)=(43​×105​×100116​×2)×(x2+1+2+1×y1+2+1+1×z2+2+3+1)=10087​x6y5z8

Multiplication of a monomial by a trinomial

Rule: In order to multiply a trinomial by a monomial, we use the following rule a×(b+c+d)=a×b+a×c+a×d ' a ' is a monomial and (b+c+d) is a trinomial.

  • Q. Multiply : 3x2−2x+2 by 3 x . Explanation : Product =3x(3x2−2x+2) =3x×3x2−3x×2x+2×3x =9x3−6x2+6x

Multiplication of a binomial by a binomial

Rule: Consider two binomials, say (a+b) and (c+d). By using the distributive property of multiplication over addition (a+b)×(c+d)=a×(c+d)+b×(c+d) ={a×c+a×d}+{b×c+b×d}=ac+ad+bc+bd

  • Q. Multiply (x+3) by (x+2) Explanation : Column method x+3 ← Multiplicand × x+2 ← Multiplier x2+3x ← First partial product i.e., x(x+3) 2x+6 ← Second partial product i.e.,2(x+3) x2+5x+6 ← Product
  • Q. Multiply (3x+5y) and (5x-7y) Solution : (3x+5y)(5x−7y) =3x×(5x−7y)+5y(5x−7y) =3x×5x−3x×7y+5y×5x−5y×7y =15x2−21xy+25yx−35y2 =15x2−21xy+25xy−35y2[∵yx=xy] =15x2+4xy−35y2

Multiplication of a binomial by a trinomial

Rule : Consider a binomial and a trinomial, say (a+b) and (c+d+e). By using the distributive property of multiplication over addition (a+b)×(c+d+e)=a×(c+d+e)+b×(c+d+e) ={a×c+a×d+a×e}+{b×c+b×d+b×e}=ac+ad+ae+bc+bd+be

  • Q. Multiply (3x2−2x+4) by (2x+5) Explanation : (2x+5)(3x2−2x+4)=2x(3x2−2x+4)+5(3x2−2x+4) =6x3−4x2+8x+15x2−10x+20 =6x3−4x2+15x2+8x−10x+20 =6x3+11x2−2x+20
  • Q. Multiply (3x2−5x+2) by (3x+5) Solution : We have, (3x2−5x+2)×(3x+5) =3x2×(3x+5)−5x(3x+5)+2×(3x+5) =(9x3+15x2)+(−15x2−25x)+(6x+10) =9x3+15x2−15x2−25x+6x+10 =9x3−19x+10
  • Q. Find the value of (5a6)×(−10ab2)×(−2.1a2b3) for a=1 and b=21​

Solution :

We have, (5a6)×(−10ab2)×(−2.1a2b3) =(5×−10×−2.1)×(a6×a×a2×b2×b3) =(5×−10×−1021​)×(a6×a×a2×b2×b3) =105a6+1+2 b2+3=105a9 b5 Putting a=1 and b=21​, we have 105a9 b5=105×(1)9×(21​)5=105×1×321​=32105​

10.0Division of Algebraic Expressions

Dividing a monomial X (say) by a monomial Y (say) means finding a monomial such that X=YZ and we write X÷Y or YX​=Z Here ' X ' is called the Dividend, ' Y ' is called the Divisor and ' Z ' is known as the Quotient.

  • While dividing variables we use the laws of exponents and powers.

Q. Divide - 72x2yz by −12xyz. Explanation :

Quotient =−12xyz−72x2yz​=(−12−72​)(xyzx2yz​)=6x

Division of a polynomial by a monomial

For dividing a polynomial in one variable by a monomial in the same variable, we perform the following steps: Step I: Obtain the polynomial (dividend) and the monomial (divisor). Step II: Arrange the terms of the dividend in descending order of their exponents. Step III: Divide each term of the polynomial by the given monomial by using the rules of division of a monomial by a monomial.

  • Q. Divide 24x3y+20x2y2−4xy by 2xy Explanation :  Quotient =2xy24x3y+20x2y2−4xy​=2xy24x3y​+2xy20x2y2​−2xy4xy​=12x2+10xy−2

Division of a polynomial by a binomial

Step I : Arrange the terms of the dividend and divisor in descending order of their exponents. Step II : Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.

Step III : Multiply the divisor by the first term of the quotient and subtract the result from the dividend to obtain the remainder.

Step IV : Consider the remainder (if any) as dividend and repeat step II to obtain the second term of the quotient. Step V : Repeat the above process till we obtain a remainder which is either zero or a polynomial of degree less than that of the divisor.

  • Change the sign of those that you are subtracting.

Division algorithm

If a polynomial is divided by another polynomial, then Dividend = Divisor × Quotient + Remainder

  • If the remainder is zero, then the divisor is a factor of the dividend.

11.0Greatest Common Factor Of Monomials

The greatest common factor (highest common factor or H.C.F.) of two or more monomials is determined as follows: (i) Find the greatest common factor of the numerical coefficients. (ii) Find the common letters appearing in the given monomials. (iii) Find the smallest power of each common letter. (iv) Multiply the numerical factor by the common letters with smallest powers obtained above.

Q. Find the greatest common factor of the monomials 6x3a2b2c,8x2ab3c3 and 12a3b2c2. Explanation : The numerical coefficients of the given monomials are 6, 8 and 12 . The greatest common factor of 6,8 and 12 is 2 . The common letters appearing in the given monomials are a,b and c . The smallest power of 'a' in the three monomials = 1 The smallest power of ' b ' in the three monomials =2 The smallest power of ' c ' in the three monomials = 1 The monomial of common letters with smallest powers =a1b2c1=ab2c Hence, the greatest common factor =2ab2c

12.0Factorisation Of Polynomials

Factorisation: Factorisation of an algebraic expression means writing the given expression as a product of its factors. These factors can be numbers, variables, or an algebraic expression.

Case-I: When each of the term contains a common monomial factor. The type is of the form 'am +bm+cm=m(a+b+c)′

Case-II : When an expression is a complete/perfect square In this case, the expression is in the form of (a) a2+2ab+b2=(a+b)2 (b) a2−2ab+b2=(a−b)2

Case-III : When an expression is the difference of two squares In this case, the expression is in the form of a2−b2=(a+b)(a−b) Process : Find the square root of each term, the sum of these roots will form one factor and their difference will form other.

13.0Algorithm

(i) Find by inspection, the greatest monomial factors which can divide each term of the expression. (ii) Find the quotient of the given expression by this monomial factor. (iii) Write this quotient in a bracket preceded by the monomial factor. Factorisation is the reverse process of multiplication.

  • Q. Factorise : 12x3y4−4x5y2 Explanation : Here 4,x3 and y2 are the greatest common factors to each term of the given expression. On dividing 12x3y4−4x5y2 by 4x3y2, we get the quotient 3y2−x2. ∴12x3y4−4x5y2=4x3y2(3y2−x2)
  • Q. Factorise : 8(p-8q) 2−6(p−8q) Solution : Here 2 and (p−8q) are the greatest common factors to both the terms. We divide the expression by 2(p−8q) and get the quotient 4(p−8q)−3. ∴8(p−8q)2−6(p−8q)=2(p−8q)[4(p−8q)−3] =2(p−8q)(4p−32q−3)
  • Q. Factorise : 25x2+4y2+20xy Explanation : 25x2+4y2+20xy=(5x)2+(2y)2+2×5x×2y=(5x+2y)2
  • Q. Factorise : 9x2+16y2−24xy Explanation : 9x2+16y2−24xy=(3x)2+(4y)2−2×3x×4y=(3x−4y)2
  • Q. Factorise : 491​x2y2−259​y2z2 Explanation : The given expression has y2 common to both the terms. We take out y2 common.
∴​491​x2y2−259​y2z2=y2(491​x2−259​z2)=y2[(71​x)2−(53​z)2]=y2(71​x+53​z)(71​x−53​z)​
  • Q. Factorise : x4−81 Explanation : x4−81=(x2)2−(9)2 =(x2+9)(x2−9) =(x2+9)(x+3)(x−3) Factorisation by regrouping terms Sometimes it is not possible to find the greatest common factor of the given set of monomials. But by regrouping the given terms, we can find the factors of the given expression.
  • Q. Factorise : (i) 1−3x−3xz+z (ii) x2−xz+xy−yz Explanation : (i) 1+z−3x(1+z)=1(1+z)−3x(1+z)=(1+z)(1−3x) (ii) x2−xz+xy−yz=(x2−xz)+(xy−yz) =x(x−z)+y(x−z)=(x+y)(x−z) Factorisation of trinomials Form-I: x2+(a+b)x+ab=(x+a)(x+b)
  • Q. Factorise: x2+7x+12 Explanation : Here, constant term =12=3×4, and 3+4=7 (coefficient of x ) ∴x2+7x+12=x2+3x+4x+12 (breaking 7x in sum of two terms, 3x and 4x ) =(x2+3x)+(4x+12)=x(x+3)+4(x+3)=(x+3)(x+4)
  • Q. Factorise: a2-a-6 Explanation : Here, constant term =−6=(−3)(2), and (−3)+(2)=−1 (coefficient of a) ∴a2−a−6=a2−3a+2a−6 =(a2−3a)+(2a−6)=a(a−3)+2(a−3)=(a−3)(a+2) Form-II: ax2+bx+c,a=1
  • Q. Factorise: 6x2+7x+2 Explanation : Here, 6×2=12=3×4 and 3+4=7 (coefficient of x ) ∴6x2+7x+2=6x2+3x+4x+2 =(6x2+3x)+(4x+2)=3x(2x+1)+2(2x+1)=(2x+1)(3x+2)
  • Q. Factorise: 1−18y−63y2 Explanation : The given expression is −63y2−18y+1. Here, (−63)×1=−63=(−21)(3) and −21+3=−18 - (coefficient of y) ∴−63y2−18y+1=−63y2−21y+3y+1=(−63y2−21y)+(3y+1) =−21y(3y+1)+1(3y+1)=(3y+1)(−21y+1)=(1+3y)(1−21y)

14.0Identities

If an algebraic expression is always equal to another expression, then it is called an identity. Square identities (i) (a+b)2=a2+2ab+b2 (ii) (a−b)2=a2−2ab+b2 (iii) (a2−b2)=(a+b)(a−b) (iv) (x+a)(x+b)=x2+(a+b)x+ab (v) (x−a)(x+b)=x2+(b−a)x−ab (vi) (x−a)(x−b)=x2−(a+b)x+ab (vii) (x+a)(x−b)=x2+(a−b)x−ab (viii) (a+b+c)2=a2+b2+c2+2ab+2bc+2ca (ix) (a−b+c)2={a+(−b)+c}2=a2+b2+c2−2ab−2bc+2ca (x) (a+b−c)2={a+b+(−c)}2=a2+b2+c2+2ab−2bc−2ca (xi) (−a+b+c)2={(−a)+b+c}2=a2+b2+c2−2ab+2bc−2ca (xii) (a−b−c)2={a+(−b)+(−c)}2=a2+b2+c2−2ab+2bc−2ca

  • Q. Solve the following using the suitable identities. (i) (x+3)(x+3) (ii) (2x+3y)2 (iii) (2a−7)(2a−7) (iv) (2x−3y)2 (v) (71)2 (vi) (99)2 (vii) 4x2−9y2 (viii) 512−492 (ix) (6x−7)(6x+7) Solution : (i) (x+3)(x+3)=(x+3)2 where a=x,b=3 [Using(a+b)2=a2+2ab+b2] =x2+2×x×3+(3)2 =x2+6x+9 (ii) (2x+3y)2 We have, (2x+3y)2=(2x)2+2×(2x)×(3y)+(3y)2[Using(a+b)2=a2+2ab+b2] =4x2+12xy+9y2 (iii) (2a−7)(2a−7) We have, (2a−7)2 [ Using (a−b)2=a2−2ab+b2] =(2a)2−2×2a×7+(7)2 =4a2−28a+49 (iv) (2x−3y)2 We have, (2x−3y)2=(2x)2−2×(2x)×(3y)+(3y)2[ Using (a−b)2=a2−2ab+b2] =4x2−12xy+9y2 (v) (71)2 We have, (70+1)2[ Using (a+b)2=a2+2ab+b2] =(70)2+2×70×1+12=4900+140+1=5041 (vi) (99)2 We have, (100−1)2 [ Using (a−b)2=a2−2ab+b2] =(100)2−2×100×1+(1)2 =10000−200+1=9800+1=9801 (vii) 4x2−9y2 We have, (2x)2−(3y)2 [ Using a2−b2=(a+b)(a−b)] =(2x+3y)(2x−3y) (viii) 512−492 We have, 512−492 [ Using a2−b2=(a+b)(a−b)] =(51+49)(51−49)=100×2=200 (ix) (6x−7)(6x+7) [Using(a−b)(a+b)=a2−b2] =(6x)2−(7)2=36x2−49
  • Q. If x+x1​=4, then find the values of (i) x2+x21​ (ii) x4+x41​ Explanation : (i) x2+x21​ We have, x+x1​=4 On squaring both sides, we get
(x+x1​)2=42⇒x2+2×x×x1​+(x1​)2=16⇒x2+2+x21​=16⇒x2+x21​=16−2[Ontransposing′2′onRHS]⇒x2+x21​=14

(ii) x4+x41​ We have, x2+x21​=14 On squaring both sides, we get

(x2+x21​)2=142⇒(x2)2+(x21​)2+2×x2×x21​=196⇒x4+x41​+2=196⇒x4+x41​=196−2[Ontransposing′2′onRHS]⇒x4+x41​=194

Equations are true for certain values of variables contained, but identities are true for all values of the contained variable.

15.0Cube Identities

(i) (a+b)3=a3+b3+3a2b+3ab2 OR a3+b3+3ab(a+b) (ii) (a−b)3=a3−b3−3a2b+3ab2 OR a3−b3−3ab(a−b) (iii) a3+b3=(a+b)(a2−ab+b2) (iv) a3−b3=(a−b)(a2+ab+b2) (v) a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca) If a+b+c=0 then, a3+b3+c3=3abc

  • Q. If a−b=8,ab=14, then find a3−b3. Solution: Since, (a−b)=8 Cubing both sides, we get, (a−b)3=83 a3−b3−3ab(a−b)=512 Substituting the values of (a−b) and ab, we get a3−b3−3×14(8)=512 ⇒a3−b3−336=512 ⇒a3−b3=512+336 ⇒a3−b3=848
  • Q. If x−x1​=2, find the value of x3−x31​. Explanation : Since x−x1​=2, Cubing both sides, we get, (x−x1​)3=23⇒x3−x31​−3⋅x⋅x1​(x−x1​)=8 Publishing \ PNCF \2024-25 \ Print Module \ SET-2 \ NCERT \ Mathematics \ 8th ⇒x3−x31​−3(x−x1​)=8 Substituting the value of (x−x1​), we get,
⇒x3−x31​−3×2=8⇒x3−x31​−6=8⇒x3−x31​=8+6⇒x3−x31​=14

Table of Contents


  • 1.0Algebraic expression
  • 2.0Important Definitions
  • 2.0.1Numerical coefficient
  • 3.0Like And Unlike Terms
  • 4.0Polynomials
  • 5.0Degree of Polynomial
  • 6.0Classification of Polynomials
  • 7.0Addition Of Algebraic Expressions
  • 8.0Subtraction Of Algebraic Expressions
  • 9.0Multiplication Of Algebraic Expressions
  • 10.0Division of Algebraic Expressions
  • 11.0Greatest Common Factor Of Monomials
  • 12.0Factorisation Of Polynomials
  • 13.0Algorithm
  • 14.0Identities
  • 15.0Cube Identities

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