If (y+z)^1/3 + (z+x)^1/3 + (x+y)^1/3 = 0. Show that (x+y+z)^3 = 9(x^3 + y^3 + z^3)

Question

Its a question regarding algebra

Answer 1

The trivial case is when x=y=z=0, so that the identity follows immediately.

 

The another way the given expression can be zero is if two of the terms are opposites

and one of them is zero

 

Without loss of generalization suppose

 y+z = -(x+y) ---> y + z = -x - y  

 

AND

z+x=0 ---> z = -x

 

y + -x = -x + -y

 

y = -y

 

y = 0

 

So y=0 and x and z are opposites

 

 

The identity holds as ( x + -x)^3 = 0 = 9*0  = 9 * (x^3 + -x^3) = 9 * (x^3 + (-x)^3)  = 9 * (x^3 + (z)^3) with y=0

 

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The only other possibility is that the sum of two of the terms in the given equation is exactly the opposite of the third term.

For example, (y+z)^(1/3) + (z+x)^(1/3) = -(x+y)^(1/3)

 

Cubing the given equation:

(y+z) + (x+z) + R(x,y,Z) = -x-y

2(x+y+z) = R(x,y,z) <---- exponents of 2/3 or less

(x+y+z) = R(x,y,z)/2

(x+y+z)^3 = R(x,y,z)^3/8 <--- exponents of 2 or less

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Then

R(x,y,z)^3/8 = 9(x^3 + y^3 + z^3)

= 9x^3 + 9y^3 + 9z^3

R(x,y,z)^3 = 72x^3 + 72y^3 + 72z^3

8(x+y+z)^3 = 72x^3 + 72y^3+ 72z^3

(x+y+z)^3 = 9x^3+9y^3+9z^2 = 9(x^3+y^3+z^2)