Question

If x=3 ,y=-2,z=0 then find the value of x3+y3+z3

Answer 1

Answer:

x = 3

y = -2

z = 0

$x^{3}+y^{3}+z^{3}=3xyz\\\\x^{3}+y^{3}+z^{3}=3(3)(-2)(0)\\\\x^{3}+y^{3}+z^{3}=0$

Answer 2

Answer:

$x = 3$

$y = 2$

$z = 0$

${x}^{3} + {y}^{3} + {z}^{3}$

${3}^{3} + {2}^{3} + {0}^{3}$

$27 + 8 + 0$

$35$