The question is below?

If x+y+z=0 then prove that (x^2+xy+y^2)^3+(y^2+yz+z^2)^3+(z^2+zx+x^2)^3=3(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)

2 Answers
Jul 11, 2018

Let

x^2+xy+y^2=a

y^2+yz+z^2=b
and

z^2+zx+x^2=c

So a-b=x^2-z^2+xy-yz

=(x-z)(x+z)+y(x-z)

=(x-z)(x+z+y)=(x-z)xx0=0

Similarly we can get

b-c=0andc-a=0

Hence a=b=c

So LHS=(x^2+xy+y^2)^3+(y^2+yz+z^2)^3+(z^2+zx+x^2)^3

=a^3+b^3+c^3=a^3+a^3+a^3=3a^3

And

RHS=3abc=3axxaxxa=3a^3

So LHS=RHS

Jul 12, 2018

Please refer to a Second Proof in Explanation.

Explanation:

Prerequisite : a=b=c rArr a^3+b^3+c^3=3abc...(star).

Let, a=x^2+xy+y^2, b=y^2+yz+z^2, &, c=z^2+zx+x^2.

Hence, (a-b)=ul(x^2-z^2)+ul(xy-yz),

=ul((x-z))(x+z)+yul((x-z)),

=(x-z)(x+z+y),

=(x-z)(0).......................................[because," Given]".

rArr a-b=0, or, a=b.

Similarly, we can show that, b=c.

Altogether, a=b=c.

:." by "(star), a^3+b^3+c^3=3abc, i.e.,

(x^2+xy+y^2)^3+(y^2+yz+z^2)^3+(z^2+zx+x^2)^3,

=3(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2).

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