1. Chapter 2 Class 9 Polynomials
  2. Serial order wise
  3. Ex 2.4
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Ex 2.4, 14 (i) - Chapter 2 Class 9 Polynomials

Last updated at Dec. 13, 2024 by Teachoo

 

 

Transcript

Ex 2.4, 14 Without actually calculating the cubes, find the value of each of the following: (-12)3 + (7)3 + (5)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 = 12, y = 7 , z = 5 (-12)3 + (7)3 + (5)3 = 3 (-12) (7)(5) + ( 12 + 7 + 5) ((-12)2 + (7)2 + (5)2 (-12)(7) (7)(5) (5)(-12) ) = 3 (-12) (7)(5) + (0) ((-12)2 + (7)2 + (5)2 (-12)(7) (7)(5) (5)(-12) = 3 (-12) (7)(5) = -1260
  1. Chapter 2 Class 9 Polynomials
  2. Serial order wise

    3. Ex 2.4

    Examples →
Examples →
Chapter 2 Class 9 Polynomials
Serial order wise

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo