Question
If x = 222, y = 223 and z = 224, then find the value of
x³ + y³ + z³ – 3xyzSolution
Given- x = 222, y = 223, z = 224 Then, as we know that – x ³ + y³ +z³ –3xyz = (x +y + z)/2 [(x-y) ² +(y-z) ² +(z-x) ²] = (222+223+224)/2 [1+1+4] = (669) ×3 = 2007 x ³ + y³ +z³ –3xyz =2007
More Algebra Questions
√3598 × √(230 ) ÷ √102= ?
15% of 2400 + (√ 484 – √ 256) = ?
(13)2 - 3127 ÷ 59 = ? x 4
6269 + 0.25 × 444 + 0.8 × 200 = ? × 15
...(53 + 480 ÷ 4)% of 20 = ?% of 70
Find the simplified value of the following expression:
62 + 122 × 5 - {272 + 162 - 422}
(15 × 225) ÷ (45 × 5) + 480 = ? + 25% of 1240
√ [? x 11 + (√ 1296)] = 16
11 × 25 + 12 × 15 + 14 × 20 + 15 = ?
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