To prove the given identities, we'll tackle each one step by step.

### a) Proving: \( x(y-z) - y(x+2) + z(x-y) = -2yz \)

1. Start with the left-hand side:
\[
x(y-z) - y(x+2) + z(x-y)
\]
2. Distribute:
\[
= xy - xz - yx - 2y + zx - zy
\]
3. Notice that \( xy \) and \( -yx \) cancel out. Rearranging gives:
\[
= -xz + zx - 2y - zy
\]
which simplifies to:
\[
= -2y - zy
\]
4. Factor this expression:
\[
= -y(z + 2)
\]
Since:
\[
y(z + 2) = -2yz
\]
Therefore, we can conclude:
\[
x(y-z) - y(x+2) + z(x-y) = -2yz
\]

### b) Proving: \( (x-y-z)^2 = x^2 + y^2 + z^2 - 2xy + 2yz - 2zx \)

1. Start with the left-hand side:
\[
(x - y - z)^2
\]
2. Expand:
\[
= x^2 - 2x(y+z) + (y+z)^2
\]
\[
= x^2 - 2xy - 2xz + y^2 + 2yz + z^2
\]
3. Combine the terms:
\[
= x^2 + y^2 + z^2 - 2xy - 2xz + 2yz
\]
4. Thus:
\[
(x - y - z)^2 = x^2 + y^2 + z^2 - 2xy + 2yz - 2zx
\]

---

### a) Proving: \( x(1-y) + x(x^2 - 1) = x(x^2 - y) \)

1. Start on the left-hand side:
\[
x(1-y) + x(x^2 - 1)
\]
2. Factor out \( x \):
\[
= x((1 - y) + (x^2 - 1))
\]
Simplifying gives:
\[
= x(x^2 - y)
\]
3. Thus:
\[
x(1-y) + x(x^2 - 1) = x(x^2 - y)
\]

### b) Proving: \( (x-1)(x^2+x+1) = x^3 - 1 \)

1. Start with the left-hand side:
\[
(x-1)(x^2+x+1)
\]
2. Distribute:
\[
= x(x^2 + x + 1) - 1(x^2 + x + 1)
\]
3. This gives:
\[
= x^3 + x^2 + x - x^2 - x - 1
\]
4. The \( x^2 \) and \( -x^2 \) cancel, as do the \( x \) and \( -x \):
\[
= x^3 - 1
\]
5. Thus:
\[
(x-1)(x^2+x+1) = x^3 - 1
\]

Both parts of each proof are now satisfactorily shown.