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State whether the following statements are true or false Give reasons to justify your answers (a) The degree of polynomial 5 x5 6 x4 8 x2 is 4 (b) The algebraic expression is a polynomial (c) The polynomial is a quadratic trinomial Using the long division method, determine the remainder when the polynomial 4 x5 2 x4 x3 4 x2 7 CBSE Class 9 Maths Lab Manual – Algebraic Identity (a b) 3 = a 3 b 3 3a 2 b 3ab 2 Objective To verify the identity (ab) 3 = a 3 b 3 3a 2 b 3ab 2 geometrically by using sets of unit cubes Prerequisite Knowledge Volume of a cube = (edge) 3 Volume of a cuboid = l x b x hVolume of cuboid = I x b x h;

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(x y)^3 identity class 9- Solution (3x– 4y) 3 is of the form Identity VII where a = 3x and b = 4y So we have, (3x – 4y) 3 = (3x) 3 – (4y) 3 – 3(3x)(4y)(3x – 4y) = 27x 3 – 64y 3 – 108x 2 y 144xy 2 Example 5 Factorize (x 3 8y 3 27z 3 – 18xyz) using standard algebraic identities Solution (x 3 8y 3 27z 3 – 18xyz)is of the form Identity Use the identity (x a) (x b) = x 2 (a b) x ab to find the following = x 2 y 2 z 2 – 6xyz 8 Question 3 Find the following squares by using the identities (i Attention reader!

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Hello Students in this video we are going to discuss our new excercise in Class 9 NCERT Maths polynomials chapter, We have already discussed till now Remaind Transcript Ex 25, 13 If x y z = 0, show that x3 y3 z3 = 3xyz We know that x3 y3 z3 3xyz = (x y z) (x2 y2 z2 xy yz zx) Putting x y z = 0, x3 y3 z3 3xyz = (0) (x2 y2 z2 xy yz zx) x3 y3 z3 3xyz = 0 x3 y3 z3 = 3xyz Hence provedSelina Concise Mathematics Part I Solutions for Class 9 Mathematics ICSE, 5 Factorisation All the solutions of Factorisation Mathematics explained in detail by experts to help students prepare for their ICSE exams

The perfect cube forms ( x y) 3 (xy)^3 (xy)3 and ( x − y) 3 ( xy)^3 (x −y)3 come up a lot in algebra We will go over how to expand them in the examples below, but you should also take some time to store these forms in memory, since you'll see them often ( x y) 3 = x 3 3 x 2 y 3 x y 2 y 3 ( x − y) 3 = x 3 − 3 x 2 y 3Class 9 Maths Polynomials Algebraic Identities Algebraic Identities Algebraic identity is an algebraic equation that is true for all values of the variables occurring in it ( x y) 2 = x2 2 xy y2 ( x – y) 2 = x2 – 2 xy y2 x2 – y2 = ( x y) ( x – y) ( x a) ( x b) = x2 ( a b) x ab (x y z) 2 = x 2 y 2 z 2 2xy 2yz 2zxClass IX Repair parts and components to include kits, assemblies, and subassemblies (repairable or nonrepairable) required for maintenance support of all equipment Class X Material to support nonmilitary programs such as agriculture and economic development (not included in Classes I through IX)

We already have an identity for (x y) 3 So, let's try to derive the identity x 3 y 3 using the identity for (x y) 3 Let's first try to understand this geometrically Let's join our cubes as shown above We arranged both cubes in such a way to convert it into a cube as shown aboveCLASSIX MATHEMATICS ASSIGNMENT2 CHAPTER – 2 POLYNOMIALS SECTIONA 1 Write the degree of the given polynomials i) ( 2x 4 )3 ii) ( t3 4 ) ( t3 9 ) 2 Write the coefficient of x4 and x in 4x3 5x4 2x2 3 3 Find the zeroes of f(z)=z2 2z 4 Find the product using suitable identities (4 5x)(45x) 5 WhatMaterials Required A set of 56 cubes each has dimensions (1 x 1 x 1) cubic unit

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Ex 25 Class 9 Maths Question 12 Verify that x 3 y 3 z 33xyz = (x y z)(x y) 2 (yz) 2 (zx) 2 Solution We have, x 3 y 3 z 3 – 3xyz = (x y z) x 2 y 2 z 2 – xy – yz – zx = (x y z)2x 2 2y 2 2z 22xy2yz 2zx = (x y z)x 2 x 2 y 2 y 2 z 2 z 22xy2yz2zx = (x y z)x 2 y 2 – 2xy y Evaluate the following using suitable identities (i) (99) 3 (ii) (102) 3 (iii) (998) 3 Solution (i) (99) 3 = (100 – 1) 3 Identity (x – y) 3 = x 3 – y 3 Polynomials Class 9 Extra Questions Very Short Answer Type Question 1 Factorise 125x 3 – 64y 3 Solution 125x 3 – 6443 = (5x) 3 – (4y) 3 By using a3 – b3 = (a – b) (a 2 ab b 2), we obtain 125x 3 – 64y 3 = (5x – 4y) (25x 2 xy 16y 2) Question 2 Find the value of (x y) 2 (x – y) 2 Solution (x y) 2 (x – y) 2 = x 2 y 2 2xy x 2 y 2 – 2xy

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Here, Right hand side = Left hand side which means that (a3) (a3) is an identity Using Activity Method In this method, the algebraic identity is verified geometrically by taking different values of a x and y Therefore, by using the identity (xy) 2 = x 22xyy 2 p 2 –10p25 = (p5) 2 (iii) 25m 2 30m9 Ans Given 25m 2 30m9 Since, 25m 2 , 30m and 9 can be substituted by (5m) 2, 2×5m×3 and 3 2 respectively we get, = (5m) 2 2×5m×3 3 2 Therefore, by using the identity (xy) 2 = x 2 2xyy 2 25m 2 30m9 = (5m3) 2 (iv) 49y 2 84yz36z 2 Ans Given 49y 2Polynomial Identities When we have a sum (difference) of two or three numbers to power of 2 or 3 and we need to remove the brackets we use polynomial identities (short multiplication formulas) (x y) 2 = x 2 2xy y 2 (x y) 2 = x 2 2xy y 2 Example 1 If x = 10, y = 5a (10 5a) 2 = 10 2 2·10·5a (5a) 2 = 100 100a 25a 2

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 Ex 25, 9 Verify (i) x3 y3 = (x y) (x2 – xy y2) Ex 25, 9 Verify (ii) x3 y3 = (x y) (x2 xy y2) LHS x3 y3 We know (x y)3 = x3 y3 3xy (x yPhone support is available MondayFriday, 900AM1000PM ET You may speak with a member of our customer support team by calling End of Conversation Have a great day!(i) To prove x 3 y 3 = (x y) (x 2 – xy y 2) Now, using identity VI we can say (x y) 3 = x 3 y 3 3xy (xy) Or, (xy) 3 – 3xy (x y) = x 3 y 3

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NCERT Class 9 Maths Lab Manual – Verify the Algebraic Identity (ab)³ = a³b³ 3a²b 3ab² OBJECTIVE To verify the algebraic identity (ab)³ = a³b³ 3a²b 3ab² Materials Required Acrylic sheets Adhesive/Adhesive tape Scissors Geometry Box Cutter Prerequisite Knowledge Concept of cuboid and its volume Concept of cube and its volume Theory CuboidRs Aggarwal 19 Solutions for Class 9 Math Chapter 3 Factorisation Of Polynomials are provided here with simple stepbystep explanations These solutions for Factorisation Of Polynomials are extremely popular among Class 9 students for Math Factorisation Of Polynomials Solutions come handy for quickly completing your homework and preparing for examsProblem Solve (x 3) (x – 3) using algebraic identities Solution By the algebraic identity, x 2 – y 2 = (x y) (x – y), we can write the given expression as;

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= (x y)(x 2 y 2 2xy x 2 xy y 2) using identity, (a b) 2 = a 2 b 2 2 ab) = (x y) (3xy) Hence, one of the factor of given polynomial is 3xy Question 18 The coefficient of x in the expansion of (x 3) 3 is (a) 1 (b) 9 (c) 18 (d) 27 Solution (d) Now, (x 3) 3 = x 3 3 3 3x (3)(x 3) using identity, (a b) 3 = a 3 b 3(x 3) (x – 3) = x 2 – 3 2 = x 2 – 9 Problem Solve (x 5) 3 using algebraic identities Solution We know, (x y) 3 = x 3 y 3 3xy(xy) Therefore, (x 5) 3 = x 3 5 3 3x5(x5) Ex 25, 11 Factorise 27 𝑥﷮3﷯ 𝑦﷮3﷯ 𝑧﷮3﷯ – 9xyz 27 𝑥﷮3﷯ 𝑦﷮3﷯ 𝑧﷮3﷯ – 9𝑥𝑦𝑧 = 3𝑥﷯﷮3﷯ 𝑦﷯﷮3﷯ 𝑧﷯﷮3﷯−9𝑥𝑦𝑧 = 3𝑥﷯﷮3﷯

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 Since x − y = 3 xy=3 x − y = 3 implies y = x − 3, y=x3, y = x − 3, substituting this into the given identity gives a x (x − 3) b x c (x − 3) 9 = 0 a x 2 (− 3 a b c) x − 3 (c − 3) = 0 \begin{aligned} ax(x3)bxc(x3)9&=0\\ ax^2(3abc)x3(c3)&=0 \end{aligned} a x (x − 3) b x c (x − 3) 9 a x 2 (− 3 a b c) x − 3 (c − 3) = 0 = 0By using standard formulae, expand the following (1 to 9) 1 (i) (2x 7y) 2 (ii) (1/2 x 2/3 y) 2 Solution (i) (2x 7y) 2 It can be written as = (2x) 2 2 × 2x × 7y (7y) 2 So we get = 4x 2 28xy 49y 2 (ii) (1/2 x 2/3 y) 2 It can be written as = (1/2 x) 2 2 × ½x 2/3y (2/3 y) 2 So we get = ¼ x 2 2/3 xy 4/9 y 2 2 (i) (3x 1/2x) 2 (ii) (3x 2 y 5z) 2We shall use the identity xy x y = x 2y 2 Here By applying in identity we get Hence the value of is (iv) The given expression is We have So we can express and in the terms of 100 as We shall use the identity xy x y = x 2y 2 Here By applying in identity

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 Transcript Example 18 Factorize 49a2 70ab 25b2 49a2 70ab 25b2 = 72 a2 70ab 52 b2 = (7a)2 70ab (5b)2 = (7a)2 (5b)2 2(7a) (5b) Using Identity (x y)2 = x2 y2 2xy where x = 7a and y = 5b = (7a 5b)2 = (7a 5b) (7a 5b) Example 18 Factorize (ii) 25/4 x2 y2/9 25/4 x2 y2/9 = 5^2/2^2 x2 y2/3^2 = (5/2 )^2 (y/3)^2 Using identity (a b) (a b) = a2 b2 In the 4th chapter of Class 9 RD Sharma Solutions students will study important identities as listed below Algebraic Identities Introduction Identity for the square of a trinomial Sum and difference of cubes Identity These books are widely used by the students who wish to score high in board examsStudents can download RD Sharma Class 9 Maths Chapter 4 exercise 44 PDF from below link Chapter 4 Algebraic Identities Rd Sharma Class 9 Solutions Maths Chapter 1 Number System Rd Sharma Class 9 Solutions Maths Chapter 2 Exponents Of Real Numbers Rd Sharma Class 9 Solutions Maths Chapter 3 Rationalisation

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For today's students, learning can happen anytime, anywhere Our mission is to improve 6th to 10th outcomes for all students and make learning more intuitive, more interesting, more personalised and more affordableAlgebraic Identities Cubic Type ,Polynomials Get topics notes, Online test, Video lectures, Doubts and Solutions for CBSE Class 9 on TopperLearningCBSE Class 9 Maths Lab Manual – Algebraic Identity (a 3 b 3) = (a b) (a 2 – ab b 2) Objective To verify the identity a 3 b 3 = (a b) (a 2 – ab b 2) geometrically by using sets of unit cubes Prerequisite Knowledge Volume of cube = (edge) 3;

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Using the above identity, the equation x 6 − y 6 can be factorised as follows x 6 Class 9 Maths Chapter 3 Unit 2 163 Qs > Related questions Factorize the following expression 8 x 3Hi everyonein this video, I tell you about " x^3y^3=(xy)(x^2xyy^2) "?channel link https//wwwyoutubecom/channel/UC7Uui8og_cIpQaH9ItVWM3Q`````Don't stop learning now Participate in the Scholorship Test for FirstSteptoDSA Course for Class 9 to 12 students My Personal Notes arrow_drop_up

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 Ex 25, 1Use suitable identities to find the following products(i) (x 4) (x 10)(x 4) (x 10)Using (x a) (x b) = x2 (a b) x ab, This video shows how to evaluate using the identity '(xy)3=x3y33x2y3xy2'To view more Educational content, please visit https//wwwyoutubecom/appuserie 9 (x y) 3 = x 3 y 3 3xy (x y) = x 3 3x 2 y 3xy 2 y 3 10 (x y) 3 = x 3 y 3 3xy (x y) = x 3 3x 2 y 3xy 2 y 3 11 x 3 y 3 z 3 3xyz = (x y z) (x 2 y 2 z 2 xy yz zx) 12 x 2 y 2 = ½ (x y) 2 (x – y) 2 13 xy = ¼ (x y) 2 (x – y) 2 14 x 2 y 2 = (x y

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(9) Verify (i) x 3 y 3 = (x y) (x 2 − xy y 2) (ii) x 3 – y 3 = (x − y) (x 2 xy y 2) using some nonzero positive integers and check by actual multiplication Can you call these as identites ?This video shows how to evaluate using the identity '(xy)3=x3y33x2y3xy2' To view more Educational content, please visit https//wwwyoutubecom/appuseriAlgebraic Identities Polynomials, Class 9, Mathematics EduRev Notes is made by best teachers of Class 9 This document is highly rated by Class 9

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CBSE Class 9 Maths Lab Manual – Algebraic Identity (a – b) 2 = a 2 – 2ab b 2 Objective To verify the identity (a – b) 2 = (a 2 – 2ab b 2) by paper cutting and pasting Prerequisite Knowledge Area of a square = (side) 2 Area of a rectangle = l x b Materials Required A white sheet of paper, glazed papers, a pair of scissorsNCERT Class 9 Maths Lab Manual – Verify the Algebraic Identity (a – b)³ = a³ – b³ – 3ab (a – b) OBJECTIVE To verify the algebraic identity (a – b)³ = a³ – b³ – 3ab (a – b) Materials Required Geometry box Acrylic sheet Scissors Adhesive/Adhesive tape CutterNCERT Class 9 Maths Lab Manual – Verify the Algebraic Identity (ab)² = a² 2abb² OBJECTIVE To verify the algebraic identity (ab)² = a² 2abb² Materials Required Drawing sheet Pencil Cellotape Coloured papers Cutter Ruler Prerequisite Knowledge Square and its area Rectangle and its area Theory A square is a quadrilateral whose all

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Exercise 25 Question 1 Use Suitable Identities To Find The Following Products X 4 X 10x 8x 10 I 3 3 3x 43x 5 Iii 2 Iv 3 2x 3 2x V Answer I By Using The Identity X Ax B X A B

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