If a = 701, b= 703 and c= 706, then find the value of a³ + b³ +c³ –3abc?
Given – a= 701, b=703, c = 706 Then as we know that a³ + b³ +c³ –3abc = (a +b + c)/2 [(a-b) ² +(b-c) ² +(c-a) ²] = (701+703+706)/2 [4+9+25] =2110 / 2 [38] =2110×19= 40090
96.03% of √225.02 × 14.98 = ? + 19.98
47.87% of 749.76 + 35.11% of 399.76 = √? + 23.15 × 20.87
2720.03 ÷ 79.98 x 39.9 = ? + 40.32
33.33% of 110.99 = 19.98% × 244.97 - √?
3.55% of 8120 – 66.66% of 540 = ? – 28% of 5500
1359.98 ÷ 30.48 × 15.12 = ? × 4.16
(0.89 3 + 1.64 3 +2.76 3 ) ÷ 5.89 = ?
12.06 × 19.02 + 12.94 × 14.87 + 152.09 = ?% of 498.98Â
(?)2 + 9.113 = 31.92 – 39.03Â
