If the value of a3+b3+c3-3abc = 10k, then find the value of k, when a = 249, b = 250 and c = 251.
225
2250
250
2500
None of these
a3+b3+c3-3abc = 0.5(a+b+c)[(a-b)2 + (b-c)2 + (c-a)2] = 0.5 x 750 x 6 = 2250
Now, 10k = 2250 ⟹ k = 225
If a+b+c=15 and a2+b2+c2=83 then find the value of a3+b3+c3−3abc