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 You are here
Ex 2.4,13
Ex 2.4, 14 (i)
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, 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