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# If a+b+c=3,a^2+b^2+c^2=13,a^3+b^3+c^3=27, then values of a b c which satisfy the three equations is

Susmita
425 Points
2 years ago
(a+b+c)2=  a2+b2+c2+2 (ab+bc+ca)
Or,9=13+2 (ab+bc+ca)
Or,(ab+ba+ca)=-2
Next,
a3+b3+c3-3abc=(a2+b2+c2-ab-bc-ca)(a+b+c)
or,27-3abc=(13+2)*9
Or,27-3abc=135
Or,3abc=27-135
Or,abc=-36
Susmita
425 Points
2 years ago
I am sorry.I missed out that it asks for values of a,b,c not abc.
Look at the 3rd equation.27 is cube of 3.So you know that one of the number is 3.You also kmow that a3+b3=0 in that case.It means that a=-b.
Look at the 2nd equation given.If c=3(as obtained from 3rd eq) then 13-9=4.Also it means that a=-b=$[\sqrt{2}]$.
Now look that these three values of a,b,c satisfies the first equation.So these are the answer.