Ex 2.4
Ex 2.4, 1 (ii)
Ex 2.4, 1 (iii) Important
Ex 2.4, 1 (iv)
Ex 2.4, 1 (v)
Ex 2.4, 2 (i) Deleted for CBSE Board 2025 Exams
Ex 2.4, 2 (ii) Important
Ex 2.4, 2 (iii)
Ex 2.4, 3 (i) Deleted for CBSE Board 2025 Exams
Ex 2.4, 3 (ii) Important
Ex 2.4, 3 (iii)
Ex 2.4, 4 (i) Deleted for CBSE Board 2025 Exams
Ex 2.4, 4 (ii)
Ex 2.4, 4 (iii) Important
Ex 2.4, 4 (iv)
Ex 2.4, 4 (v)
Ex 2.4, 4 (vi)
Ex 2.4, 5 (i) Deleted for CBSE Board 2025 Exams
Ex 2.4, 5 (ii) Important
Ex 2.4, 6 (i)
Ex 2.4, 6 (ii)
Ex 2.4, 6 (iii)
Ex 2.4, 6 (iv) Important
Ex 2.4, 7 (i) Deleted for CBSE Board 2025 Exams
Ex 2.4, 7 (ii)
Ex 2.4, 7 (iii) Important
Ex 2.4, 8 (i)
Ex 2.4, 8 (ii)
Ex 2.4, 8 (iii) Important
Ex 2.4, 8 (iv) Important
Ex 2.4, 8 (v)
Ex 2.4, 9 (i)
Ex 2.4, 9 (ii)
Ex 2.4, 10 (i) Important Deleted for CBSE Board 2025 Exams
Ex 2.4, 10 (ii)
Ex 2.4, 11 Deleted for CBSE Board 2025 Exams
Ex 2.4,12 Important
Ex 2.4,13
Ex 2.4, 14 (i) You are here
Ex 2.4, 14 (ii) Important
Ex 2.4, 15 (i)
Ex 2.4, 15 (ii) Important
Ex 2.4, 16 (i) Deleted for CBSE Board 2025 Exams
Ex 2.4, 16 (ii) Important
Ex 2.4
Last updated at April 16, 2024 by Teachoo
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