1. Chapter 2 Class 9 Polynomials
2. Serial order wise

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Ex 2.5, 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 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
Serial order wise