
Ex 2.5
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 You are here
Ex 2.5,13 Deleted for CBSE Board 2022 Exams
Ex 2.5, 14 (i) Deleted for CBSE Board 2022 Exams
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
Last updated at Aug. 25, 2021 by Teachoo
Ex 2.5, 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