Question

(x+y+z)(x+yw+zw^2)(x+yw^2+zw)=

Answer 1

1. multiply first two terms

we get,

x*2+wxy+w*2xz+xy+w*2y*2+w*2yz+xz+wyz+w*2z*2.

x*2+xy[1+w]+xz[1+w*2]+yz[w+w*2]+y*2w+z*2w*2.

x*2+xy(-w*2) +xz(-w) +yz(-1)+y*2w+z*2w*2.

2. Now, multiply with third term,

we get, x*3+xy*2w+xz*2w*2-x*2yw*2-x*2zw-xyz+x*2yw*2+y*3w*3+yz*2w*4-xy*2w*4-xyzw*3-y*2w*2+x*2zw+y*2zw*2+z*3w*3-xyzw*3-xz*2w*2-yz*2w.

we know that,

a. 1+w+w*2=0

b. w*3 = 1

c. w*4=w*3.w=(1).w= w

# w*4=w

Now, we get

x*3+y*3+z*3+xy*2(w*4-w) +xz*2(w*2-w*2)-x*2y(w*2-w*2)-x*2z(w-w) -xyz(1+w*3+w*3)+yz*2(w*4-w) -y*2z(w*2-w*2).

Here, all terms will cancel and finally after Cancelling we get,

x*3+y*3+z*3-xyz(1+w*3+w*3)

x*3+y*3+z*3-xyz(1+1+1)

x*3+y*3+z*3-xyz(3)

x*3+y*3+z*3-3xyz

Finally, [x+y+z] [x+wy+w*2z] [x+w*2y+wz] =

x*3+y*3+z*3-3xyz.....