If a+b+c =6 and ab+bc+ac=11, find the value of a^3+b^3+c^3-3abc.
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Answered by
1
Answer:
the solution is In the above.
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Answered by
1
Answer:
18
Step-by-step explanation:
(a+b+c)^2=ab + 2bc + 2ca
36=a^2 + b^2 + c^2 + 2(ab+bc+ca)
36=a^2 + b^2 + c^2 + 2(11)
36-22=a^2 + b^2 + c^2
14=a^2 + b^2 + c^2
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
=(6)(14-11)
=6*3=18
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