Question

$x^3 + y^3 + z^3 = 3abc$
a, b, c - length of a triangle ABC
What kind of triangle is it?

Answer 1

Answer:

equilateral triangle

Step-by-step explanation:

$x^3+y^3+z^3-3abc=0\\(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0\\(x+y+z)*\frac{1}{2} *[(x-y)^2+(y-z)^2+(z-x)^2]=0\\x+y+z>0\\So,(x-y)^2+(y-z)^2+(z-x)^2=0\\(x-y)^2=0, (y-z)^2=0, (z-x)^2=0\\x=y, y=z, z=x\\x=y=z$

Answer 2

Answer:

???

Step-by-step explanation:

YOUR QUESTION IS WRONG