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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).

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).

Grade:

2 Answers

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).

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:

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

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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Sagar Singh

B.Tech, IIT Delhi

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