1. Chapter 2 Class 9 Polynomials (Term 2)
  2. Serial order wise
  3. Ex 2.5

Ex 2.5, 14 (ii) - Chapter 2 Class 9 Polynomials (Term 2)

Last updated at Aug. 25, 2021 by

Ex 2.5,14 - Chapter 2 Class 9 Polynomials - Part 2

Advertisement

Advertisement

  1. Chapter 2 Class 9 Polynomials (Term 2)
  2. Serial order wise

    3. Ex 2.5

    Examples →

Transcript

Ex 2.5, 14 Without actually calculating the cubes, find the value of each of the following: (ii) (28)3 + (-15)3 + (-13)3 We know that x3 + y3 + z3 3xyz = (x + y + z) (x2 + y2 + z2 xy yz zx) So, x3 + y3 + z3 = 3xyz + (x + y + z) (x2 + y2 + z2 xy yz zx) Putting x = 28, y = 15 , z = 13 (28)3 + ( 15)3 + ( 13)3 = 3 (28) ( 15) ( 13) + (28 15 13) ((28)2 +(-15)2+(-13)2 +(28)(-15)+(-15)(-13)+(-13)(28)) = 3(28)( 15)( 13) + 0 = 16380

Chapter 2 Class 9 Polynomials (Term 2)
Serial order wise

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.