suryakanth AskiitiansExpert-IITB
Last Activity: 14 Years ago
If x+ y+ z=0
Prove that:
(x²+xy+y²)³+ (y²+yz+z²)³+ (z²+zx+x²)³ = 3(x²+xy+y²). (y²+yz+z²) (z²+zx+x²).
Answer:
Let a = (x²+xy+y²)
b= (y²+yz+z²)
c= (z²+zx+x²)
This equation reduces to proving that a3+b3+c3=3abc
This is possible if:
We realize by simple substitution(like taking x,y,z = (-1,0,1),(-2,0,2)) that a+b+c is not 0 all the times
Now considering a=b
If and only if
(x²+xy+y²) = (y²+yz+z²)
If and only if
x²+xy = yz+z²
i.e., x(x+y)=z(y+z)
Using the fact that x+y+z=0, we see this is nothing but,
x(-z)=z(-x)
Hence a=b=c
=> a3+b3+c3=3abc
which is
(x²+xy+y²)³+ (y²+yz+z²)³+ (z²+zx+x²)³ = 3(x²+xy+y²). (y²+yz+z²) (z²+zx+x²).
Hence proved.
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