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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)#

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Answer 1 · 2018-07-11T10:39:58

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#

Answer 2 · 2018-07-12T06:14:00

Please refer to a Second Proof in 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)#.

#color(red)("Enjoy Maths!")#