Factor x^2(x - y) + (y - x) completely.

Step-by-step solution

1. 1

   Notice the near-match between the two brackets. The expression contains $(x-y)$ and $(y-x)$ — not identical, but opposites. This is the standard obstacle in factoring by grouping, and it is fixed by a single sign flip rather than by expanding everything.

2. 2

   Flip the second bracket. Since

   $y-x=-(x-y)$

   the expression becomes

   $x^2(x-y)-(x-y)$

   Both terms now share the identical factor $(x-y)$.

3. 3

   Factor out the common bracket. Writing the second term as $1\cdot(x-y)$ makes the coefficient explicit:

   $x^2(x-y)-1(x-y)=(x-y)\left(x^2-1\right)$

4. 4

   Recognise the difference of squares. $x^2-1=x^2-1^2$, so by $a^2-b^2=(a-b)(a+b)$:

   $x^2-1=(x-1)(x+1)$

   Leaving the answer as $(x-y)(x^2-1)$ would be incomplete factoring.

5. 5

   Write the complete factorisation.

   $x^2(x-y)+(y-x)=(x-y)(x-1)(x+1)$

   Three linear factors, so the expression vanishes exactly when $x=y$, $x=1$, or $x=-1$.

6. 6

   Verify numerically and by expanding. Testing $60$ random pairs $(x,y)$ in $[-4,4]$ gives agreement to machine precision at every point ✓. As a spot check at $x=3,y=1$: the original is $9(2)+(1-3)=18-2=16$, and the factored form is $(2)(2)(4)=16$ ✓.