Question

x,y,z are the three natural number whose sum is divisible by 6.prove that x^3+y^3+z^3 is divisible by 6​

Answer 1

Answer:

Step-by-step explanation:

given :

        (x+y+z)%6=0

to prove:

       (x^3+y^3+z^3)%6=0

solution:

(x^3+y^3+z^3) = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)

Divide both side by 6,

(x^3+y^3+z^3)%6 = (x + y + z)%6 (x2 + y2 + z2 – xy – yz – zx)  

{given: (x+y+z)%6=0}

(x^3+y^3+z^3)%6=0 hence prove