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# 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).

105 Points
10 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.

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.

SAGAR SINGH - IIT DELHI
879 Points
10 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....

All the best.

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