(x-y)^3=x^3-y^3-3xy(x-y) 131676

(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) = x 3 125 15x(x5) = x 3 125 15x 2Formula ${(xy)}^3$ $\,=\,$ $x^3y^33xy(xy)$ Introduction $xy$ is a binomial in which $x$ and $y$ are two terms In mathematics, the cube of sum of two terms isThe cube of a binomial is defined as the multiplication of a binomial 3 times to itself We know that cube of any number 'y' is expressed as y × y × y or y 3, known as a cube numberTherefore, given a binomial which is an algebraic expression consisting of 2 terms ie, a b, the cube of this binomial can be either expressed as (a b) × (a b) × (a b) or (a b) 3

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(x-y)^3=x^3-y^3-3xy(x-y)

(x-y)^3=x^3-y^3-3xy(x-y)- = (4a – 3b) 3 ∵ (x – y) 3 = x 3 – y 3 – 3xy(x – y) = (40 – 3b) (4a – 3b) (4a – 3b) Question 3 What are the possible expressions for the dimensions of a cuboid whose volume is given below ?Devkit is a vibrant, endtoend website template pack for open source software projects, SaaS and API services

X3 Y3 Identity

NCERT Solutions for Class 9 Maths Chapter 2 Polynomial Ex 25 given here are prepared by the subject experts Visit now to download solutions in PDF for freeSolution By the algebraic identity, x 2 – y 2 = (x y) (x – y), we can write the given expression as;Here, exponent of every variable is a whole number, but x 10 y 3 t 50 is a polynomial in x, y and t, ie, in three variables So, it is not a polynomial in one variable Ex 21 Class 9 Maths Question 2

Volume = 12ky 2 8ky – k Solution We have, volume = 12ky 2 8ky – k = 4k(3y 2 2y – 5) = 4k(3y 2 5y – 3y – 5)

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