Question

The expression x(xy + xz) + y(xy + yz) + z(xz + yz) is even if and only if:
1] x is even. 2) xy is odd. 3) xyz is odd. 4] Any of these.​

Answer 1

Given :  x(xy + xz) + y(xy + yz) + z(xz + yz)  is even  

To Find : expression is even  if and only if:

1] x is even. 2) xy is odd. 3) xyz is odd. 4] Any of these.​

Solution:

x(xy + xz) + y(xy + yz) + z(xz + yz)

= x(xy)  + x(xz) + y(xy)  + y(yz) + z(xz)  + z(yz)

= xy(x + y)  + xz(x + z)  + yz(y + z)

Each term is of form   ab(a + b)

case 1 : a  odd  , b  odd

=> odd * odd ( odd + odd)  = odd ( even)  = even

case 2 : a  odd  , b  even

=> odd *  even ( odd +  even)  = even ( odd )  = even

case 3 : a  even , b  odd  

=> even * odd ( even + odd)  = even ( odd )  = even

case 4 : a  even  , b  even

=> even * even ( even + even)  = even  ( even)  = even

Hence whatever the combination is , each term is even

even + even + even = even

=> xy(x + y)  + xz(x + z)  + yz(y + z) is always even

Hence x(xy + xz) + y(xy + yz) + z(xz + yz)  is always even  independent of x , y & z being odd/even

 x is even.

xy is odd   if  x or/both y are odd

xyz is odd.   if any one of x/y/z is odd or all x , y , z are odd

Hence Any of these.​ is correct

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