If a+b+c=11, and ab+bc+ca=25 then find the value of a^3+b^3+c^3-3abc by using suitable identity.
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Use the identity (a+b)^3 = a^3+b^3+3 a b (a+b).
[(a+b) + c]^3 = (a+b)^3 + c^3 + 3 (a+b) c (a+b+c)
So 11^3 = a^3+b^3+3ab(a+b) + c^3 + 3 (a+b)c *11
1331 = a^3 + b^3 + c^3 + 3 a b (11 - c) + 33 (ac+bc)
so a^3+b^3+c^3-3abc = 1331 - 33 (ab+bc+ca)
= 1331 - 33*25
= 506
[(a+b) + c]^3 = (a+b)^3 + c^3 + 3 (a+b) c (a+b+c)
So 11^3 = a^3+b^3+3ab(a+b) + c^3 + 3 (a+b)c *11
1331 = a^3 + b^3 + c^3 + 3 a b (11 - c) + 33 (ac+bc)
so a^3+b^3+c^3-3abc = 1331 - 33 (ab+bc+ca)
= 1331 - 33*25
= 506
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