Ex 2.4, 14 (ii) - Chapter 2 Class 9 Polynomials

Last updated at Dec. 16, 2024 by

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Ex 2.4, 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

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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