thanks. how about this?
factor completely
y^2(x^4-z^4) - x^2(y^4-z^4) +z^2(y^4-x^4)
i really have a hard time in factoring long expressions. is there any technique? thanks
You started with $\displaystyle y^2(x^4\, -\, z^4)\, -\, x^2(y^4\, -\, z^4)\, +\, z^2(y^4\, -\, x^4)$
In the above, you multiplied everything out to get:
. . . . .$\displaystyle x^4 y^2\, -\, y^2 z^4\, -\, x^2 y^4\, +\, x^2 z^4\, +\, y^4 z^2\, -\, x^4 z^2$
There appear to be some steps missing between the above and your next step:
I think you regrouped and then factored out of pairs of terms, as:
. . . . .$\displaystyle x^4 y^2\, -\, x^2 y^4\, +\, x^2 z^4\, -\, y^2 z^4\, -\, x^4 z^2\, +\, y^4 z^2$
. . . . .$\displaystyle x^2 y^2 (x^2\, -\, y^2)\, +\, z^4 (x^2\, -\, y^2)\, -\, z^2 (x^4\, -\, y^4)$
Then you factored the difference of squares in the last set of parentheses above, and factored "$\displaystyle x^2\, -\, y^2$" out front:
. . . . .$\displaystyle x^2 y^2 (x^2\, -\, y^2)\, +\, z^4 (x^2\, -\, y^2)\, -\, z^2 \left[(x^2\, -\, y^2)(x^2\, +\, y^2)\right]$
. . . . .$\displaystyle (x^2\, -\, y^2)\left[x^2 y^2\, +\, z^4\, -\, z^2(x^2\, +\, y^2)\right]$
. . . . .$\displaystyle (x^2\, -\, y^2)(x^2 y^2\, +\, z^4\, -\, x^2 z^2\, -\, y^2 z^2)$
You then paired and factored inside the second parentheses:
. . . . .$\displaystyle (x^2\, -\, y^2)(x^2 y^2\, -\, x^2 z^2\, -\, y^2 z^2\, +\, z^4)$
. . . . .$\displaystyle (x^2\, -\, y^2)\left[x^2(y^2\, -\, z^2)\, -\, z^2(y^2\, -\, z^2)\right]$
. . . . .$\displaystyle (x^2\, -\, y^2)\left[(y^2\, -\, z^2)(x^2\, -\, z^2)\right]$
Then you factored the differences of squares:
. . . . .$\displaystyle (x\, -\, y)(x\, +\, y)\left[(y\, -\, z)(y\, +\, z)(x\, -\, z)(x\, +\, z)\right]$
This wasn't the factored form I'd expected, but it looks to be correct. Good job!