 # If x+ y+ z=0 then prove that (x²+xy+y²)³+ (y²+yz+z²)³+ (z²+zx+x²)³ = 3(x²+xy+y²). (y²+yz+z²) (z²+zx+x²). suryakanth AskiitiansExpert-IITB
105 Points
12 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).

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:

• a+b+c = 0
• or a=b=c

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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12 years ago

Dear ajit,

Take LHS

(x²+xy+y²)³+ (y²+yz+z²)³+ (z²+zx+x²)³

Multiply by [(x-y)((y-z)(z-x)]^3 and simplify we will get RHS....

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All the best.

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