Question

if x +y+z=0 show that (x3+y3+z3)=3xyz​

Answer 1

Answer:

We have x

3

+y

3

+z

3

−3xyz

=(x+y+z)(x

2

+y

2

+z

2

−xy−yz−zx)

0=(x+y+z)(x

2

+y

2

+z

2

−xy−yz−zx)

So either x+y+z=0 (or)

x

2

+y

2

+z

2

−xy−yz−zx=0

x

2

+y

2

+z

2

−xy−yz−zx=0

2x

2

+2y

2

+2z

2

−2xy−2yz−2zx=0

(x

2

−2xy+y

2

)+(y

2

−2yz+z

2

)+(z

2

−2zx+x

2

)=0

(x−y)

2

+(y−z)

2

+(z−x)

2

=0

⇒x=y=z

Therefore, either x+y+z=0 or x=y=z.

Step-by-step explanation:

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