hey is the second one a"cube"?

yes

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hey is the second one a"cube"?

yes

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ok

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You didn't specify if a,b,c are all distinct. Firste note that:

a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc).

By assumption a^3+b^3+c^3=3abc so the left hand side is 0.

Therefore (a+b+c)(a^2+b^2+c^2-ab-ac-bc) = 0. So either a+b+c=0 or a^2+b^2+c^2-ab-ac-bc=0.

Now suppose a^2+b^2+c^2 - ab-ac-bc =0. You can re-write this as:

(1/2) ( (a-b)^2 + (b-c)^2 + (c-a)^2 )=0

Therefore (a-b)^2 + (b-c)^2 + (c-a)^2 =0 so a=b=c.

So either a+b+c=0 or a=b=c. (So if a,b,c are all distinct then a+b+c=0).

a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc).

By assumption a^3+b^3+c^3=3abc so the left hand side is 0.

Therefore (a+b+c)(a^2+b^2+c^2-ab-ac-bc) = 0. So either a+b+c=0 or a^2+b^2+c^2-ab-ac-bc=0.

Now suppose a^2+b^2+c^2 - ab-ac-bc =0. You can re-write this as:

(1/2) ( (a-b)^2 + (b-c)^2 + (c-a)^2 )=0

Therefore (a-b)^2 + (b-c)^2 + (c-a)^2 =0 so a=b=c.

So either a+b+c=0 or a=b=c. (So if a,b,c are all distinct then a+b+c=0).

So if a+b+c = 0

then a^3 + b^3 + c^3 - 3abc = 0

So a^3 + b^3 + c^3 = 3abc

Hope it helps and pls mark as the best answer. :-)