3a^2bc+6ab^2c+9abc^2 explane
Answers
There are tons of similar questions, and usually the shortest answer comes from the use of Vietta theorem.
What do I mean: let a, b, c be the roots of the polynomial p(x)=x3+ux2+vx+w
Now we know that p(a)=p(b)=p(c)=0
Sum those equations up and get:
a3+b3+c3+u(a2+b2+c2)+v(a+b+c)+3w=0
But you know that
u=−(a+b+c)
v=ab+bc+ac
w=−abc
Thus you equation turns into:
(a+b+c)(a2+b2+c2−(ab+bc+ac))=0
So we have 2 choices:
a+b+c=0
a2+b2+c2=ab+bc+ac
The right part of the second equation is the scalar product of vectors r1=(a,b,c) and r2=(b,c,a). Thus ab+bc+aca2+b2+c2=(r1,r2)|r1||r2|=cos(r1;r2)=1. As both vectors have the same length this may occur only if r1=r2 i.e. a=b=c.
As for our case a=b=c is not a solution the only one that is left is a+b+c=0
P.S.: The similar exercises appears nearly once a month on Quora. I hope people search a bit when they ask a question.
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(a+b+c)=0
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If a3+b3+c3–3abc=0 then either a+b+c=0or a2+b2+c2-ab-bc-ca=0
if we multiply both sides by 2 it becomes,
(a-b)2+(b-c)2+(c-a)2=0 as all three terms are greater or equal to 0
then a=b=c
therefore,a3=b3=c3=3abc if a+b+c=0 or a=b=c.
this means a+b+c has to be zero
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0
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1 by logic
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Ok let's go ok let's go with the formula
Here
a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
Since ,
RHS =0
Hence,
a+b+c =0