If ab+bc+ca=10 and a^2+b^2+c^2=44 then find a^3+b^3+c^3-3abc
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(a+b+c)³=(a³+b³+c³)+3[(a+b+c)(ab+ac+bc)−abc]
(a+b+c)³ = (a³+b³+c³) + 3(10)(a+b+c) -3abc
(a³+b³+c³) - 3abc = (a+b+c)³ - 30(a+b+c)
(a³+b³+c³) - 3abc = (a+b+c) [ (a+b+c)^2 - 30 ]
(a³+b³+c³) - 3abc = (a+b+c) [ 44 +10 -30 ]
I can do it only till here. Sorry.
himantikagupta:
Wrong answer
Answered by
5
Answer:
(a+b+c)³=(a³+b³+c³)+3[(a+b+c)(ab+ac+bc)−abc]
(a+b+c)³ = (a³+b³+c³) + 3(10)(a+b+c) -3abc
(a³+b³+c³) - 3abc = (a+b+c)³ - 30(a+b+c)
(a³+b³+c³) - 3abc = (a+b+c) [ (a+b+c)^2 - 30 ]
(a³+b³+c³) - 3abc = (a+b+c) [ 44 +10 -30 ]
I can do it only till here. Sorry
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