1. Chapter 2 Class 9 Polynomials
  2. Serial order wise
  3. Ex 2.4
Check sibling questions

Ex 2.4,12 - Chapter 2 Class 9 Polynomials

Last updated at Dec. 13, 2024 by Teachoo

Transcript

Ex 2.4, 12 Verify that x3 + y3 + z3 – 3xyz = 1/2 (x + y + z)[(x – y)2 + (y – z)2 + (z – x)2] Solving R.H.S 1/2 (x + y + z)[(x – y)2 + (y – z)2 + (z – x)2] Using (a - b)2 = a2 + b2 - 2ab = 1/2 (x + y + z) [ (x2 + y2 – 2xy) + (y2 + z2 – 2yz) + (z2 + x2 – 2zx)] = 1/2 (x + y + z) [2x2 + 2y2 + 2z2 – 2xy– 2yz – 2zx] = 1/2 (x + y + z) 2 [x2 + y2 + z2 – xy – yz – zx] = (x + y + z) [x2 + y2 + z2 – xy – yz – zx] We know x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx) = x3 + y3 + z3 – 3xyz = L.H.S
  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