Simplify (x^2 - 2yz + z^2) + (3yz - z^2 + 5x^2).

Step-by-step solution

1. 1

   Drop the brackets — addition permits it directly. Because the two groups are added (not subtracted), every sign inside survives unchanged:

   $x^{2}-2yz+z^{2}+3yz-z^{2}+5x^{2}$

   Had there been a minus sign in front of the second bracket, every one of its signs would have had to flip — the single most common error in this type of question.

2. 2

   Group the like terms. Three distinct variable parts appear: $x^{2}$, $yz$ and $z^{2}$. Collect each:

   $\left(x^{2}+5x^{2}\right)+\left(-2yz+3yz\right)+\left(z^{2}-z^{2}\right)$

3. 3

   Combine each group.

   $x^{2}+5x^{2}=6x^{2},\qquad-2yz+3yz=yz,\qquad z^{2}-z^{2}=0$

   Note $-2yz+3yz=+1yz$, written simply as $yz$, and the $z^{2}$ terms annihilate completely so nothing of them remains.

4. 4

   Write the result.

   $6x^{2}+yz$

   Only two terms survive from the original six. They cannot be combined further because $x^{2}$ and $yz$ involve different variables.

5. 5

   Verify numerically. At $x=1$, $y=2$, $z=3$: the first bracket is $1-12+9=-2$ and the second is $18-9+5=14$, summing to $12$; the answer gives $6(1)+2(3)=12$ ✓. The identity was confirmed at all $1728$ integer triples with $-6\le x,y,z\le5$ ✓.