If a = 901, b= 903, and c= 906, then find the value of a³ + b³ +c³ –3abc?
Given – a= 901, b=903, c = 906 Then as we know that a³ + b³ +c³ –3abc = (a +b + c)/2 [(a-b) ² +(b-c) ² +(c-a) ²] = (901+903+906)/2 [4+9+25] =2710 / 2 [38] =2710×19= 51490
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