When x+y+z =16 and xy + yz+zx =78, find the value of $$x^3 +y^3 +z^3-3xyz.$$​​

Correct option is A

Given:

​$$x+y+z=16$$ 

$$xy+yz+zx=78$$ 

$$x^3 + y^3 + z^3 - 3xyz$$ = ?

Formula Used:

​$$x^3 + y^3 + z^3 - 3xyz = (x + y + z)\left( (x + y + z)^2 - 3(xy + yz + zx) \right)$$ 

Solution:

$$x^3 + y^3 + z^3 - 3xyz = (x + y + z)\left( (x + y + z)^2 - 3(xy + yz + zx) \right)$$ 

​Now, let's substitute given values into the identity

​$$x^3 + y^3 + z^3 - 3xyz = (16)\left( (16)^2 - 3(78) \right)$$ 

$$x^3 + y^3 + z^3 - 3xyz = 16 \left( 256 - 234 \right)$$ 

$$x^3 + y^3 + z^3 - 3xyz = 16 \times 22 = 352$$ 

Thus $$x^3 + y^3 + z^3 - 3xyz$$ is 352.​