Factor 6x(x − y)² + 3(y − x)(y − x)².

Step-by-step solution

1. 1

   Collapse the second term into a single power. The product $(y-x)(y-x)^2$ is just $(y-x)^3$:

   $6x(x-y)^2 + 3(y-x)^3$

   Writing it as one power makes the parity of the exponent visible, which is what the next step depends on.

2. 2

   Align the two binomials using the sign rule. Since $y - x = -(x-y)$, an odd power picks up a minus sign while an even power does not:

   $(y-x)^3 = \left[-(x-y)\right]^3 = -(x-y)^3, \qquad (y-x)^2 = (x-y)^2$

   This is the single step where most sign errors happen: only odd exponents flip.

3. 3

   Rewrite the whole expression in terms of (x − y).

   $6x(x-y)^2 + 3(y-x)^3 = 6x(x-y)^2 - 3(x-y)^3$

   Now both terms share the same binomial base, so a common factor can be extracted.

4. 4

   Take out the greatest common factor. The numerical GCF of $6$ and $3$ is $3$, and the lower power of the binomial is $(x-y)^2$:

   $6x(x-y)^2 - 3(x-y)^3 = 3(x-y)^2\Big[2x - (x-y)\Big]$

5. 5

   Simplify the remaining bracket.

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

   $\Longrightarrow \quad 6x(x-y)^2 + 3(y-x)(y-x)^2 = 3(x-y)^2(x+y)$

6. 6

   Check with test values. At $(x,y) = (3,5)$ the original is $6(3)(4) + 3(2)(4) = 72 + 24 = 96$, and the factored form gives $3(4)(8) = 96$. The same agreement holds at $(-2, 7)$ and $\left(\tfrac12, \tfrac13\right)$ in exact fraction arithmetic. Neither $(x-y)^2$ nor $x+y$ factors further, so this is complete.