Question
If a = 701, b= 703 and c= 706, then find the value of
a³ + b³ +c³ –3abc?Solution
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
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