Last updated at Aug. 25, 2021 by Teachoo

Transcript

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

Ex 2.5, 1 (i)

Ex 2.5, 1 (ii)

Ex 2.5, 1 (iii) Important

Ex 2.5, 1 (iv)

Ex 2.5, 1 (v)

Ex 2.5, 2 (i)

Ex 2.5, 2 (ii) Important

Ex 2.5, 2 (iii)

Ex 2.5, 3 (i)

Ex 2.5, 3 (ii) Important

Ex 2.5, 3 (iii)

Ex 2.5, 4 (i)

Ex 2.5, 4 (ii)

Ex 2.5, 4 (iii) Important

Ex 2.5, 4 (iv)

Ex 2.5, 4 (v)

Ex 2.5, 4 (vi)

Ex 2.5, 5 (i)

Ex 2.5, 5 (ii) Important

Ex 2.5, 6 (i)

Ex 2.5, 6 (ii)

Ex 2.5, 6 (iii)

Ex 2.5, 6 (iv) Important

Ex 2.5, 7 (i)

Ex 2.5, 7 (ii)

Ex 2.5, 7 (iii) Important

Ex 2.5, 8 (i)

Ex 2.5, 8 (ii)

Ex 2.5, 8 (iii) Important

Ex 2.5, 8 (iv) Important

Ex 2.5, 8 (v)

Ex 2.5, 9 (i)

Ex 2.5, 9 (ii)

Ex 2.5, 10 (i) Important

Ex 2.5, 10 (ii)

Ex 2.5, 11

Ex 2.5,12 Important Deleted for CBSE Board 2022 Exams

Ex 2.5,13 Deleted for CBSE Board 2022 Exams

Ex 2.5, 14 (i) Deleted for CBSE Board 2022 Exams You are here

Ex 2.5, 14 (ii) Important Deleted for CBSE Board 2022 Exams

Ex 2.5, 15 (i)

Ex 2.5, 15 (ii) Important

Ex 2.5, 16 (i)

Ex 2.5, 16 (ii) Important

About the Author

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.