If a+b+c=6 and ab+bc+ca=11, find the value of a3+b3+c3-3abc.

Given, a + b + c = 6 and ab + bc + ca = 11

Now, a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca)

Again, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

⇒ a2 + b2 + c2 = (a + b + c)2 - 2 (ab + bc + ca)

⇒ a2 + b2 + c2 = (6)2 - 2 (11) = 36 - 22 = 14

⇒a3 + b3 + c3 - 3abc = (a + b + c) [a2 + b2 + c2 - (ab + bc + ca)]

⇒a3 + b3 + c3 - 3abc = (6) [14 - (11)]

⇒a3 + b3 + c3 - 3abc = (6) [14 - (11)] = 6(3) = 18

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