Question

If x^3+ y^3 + z^3 = 3 xyz, then x+y+z​

Answer 1

Answer:

We know that:

x

3

+y

3

+z

3

−3xyz=(x+y+z)(x

2

+y

2

+z

2

−xy−yz−zx).

Let us consider, (x+y+z)=0.

x

3

+y

3

+z

3

−3xyz=(0)(x

2

+y

2

+z

2

−xy−yz−zx)

⇒x

3

+y

3

+z

3

−3xyz=0

⇒x

3

+y

3

+z

3

=3xyz.

Then, if x

3

+y

3

+z

3

=3xyz, then (x+y+z)=0.

Answer 2

Answer:

We know that:

$x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx).$

Let us consider,

$(x+y+z)=0$

$x3+y3+z3−3xyz=(0)(x2+y2+z2−xy−yz−zx)</p><p></p><p>⇒x3+y3+z3−3xyz=0</p><p></p><p>⇒x3+y3+z3=3xyz</p><p></p><p>$

Then, if

$x3+y3+z3=3xyz$

,then (x+y+z)=0.

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