Question
Simplify:
$(x^2 + y^2 - z^2)^2 - (x^2 - y^2 + z^2)^2$
$(x^2 + y^2 - z^2)^2 - (x^2 - y^2 + z^2)^2$
✓
Answer
We have $\left(x^2+y^2-z^2\right)^2-\left(x^2-y^2+z^2\right)^2$
Using formula $(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x$,
we get $\left(x^2+y^2-z^2\right)^2-\left(x^2-y^2+z^2\right)^2$
$=(\text{x}^2)^2+(\text{y}^2)^2+(-\text{z}^2)^2+2(\text{x}^2)(\text{y}^2)+2(\text{y}^2)(-\text{z}^2)+2(-\text{z}^2)(\text{x}^2)\\-\Big[(\text{x}^2)^2+(-\text{y}^2)^2+(\text{z}^2)^2+2(\text{x}^2)(-\text{y}^2)+2(-\text{y}^2)(\text{z}^2)+2(\text{z}^2)(\text{x}^2)\Big]$
$=\text{x}^4+\text{y}^4+\text{z}^4+2\text{x}^2\text{y}^2-2\text{y}^2\text{z}^2-2\text{z}^2\text{x}^2\\-\big[\text{x}^4+\text{y}^4+\text{z}^4-2\text{x}^2\text{y}^2-2\text{y}^2\text{z}^2+2\text{z}^2\text{x}^2\big]$
By canceling the opposite terms,
we get$\big(\text{x}^2 +\text{y}^2 -\text{z}^2\big)^2 -\big(\text{x}^2 -\text{y}^2+\text{z}^2)^2=\not\text{x}^4+\not\text{y}^4+\not\text{z}^4\\+2\text{x}^2\text{y}^2-2\text{y}^2\text{z}^2-2\text{z}^2\text{x}^2\not\text{x}^4-\not\text{y}^4-\not\text{z}^4+2\text{x}^2\text{y}^2+2\text{y}^2\text{z}^2-2\text{z}^2\text{x}^2$
$=4\text{x}^2\text{y}^2-4\text{z}^2\text{x}^2$
Talking $4x^2$ as common factor we get$\big(\text{x}^2 +\text{y}^2 -\text{z}^2\big)^2 -\big(\text{x}^2 -\text{y}^2+\text{z}^2)^2=4\text{x}^2(\text{y}^2-\text{z}^2)$
Hence the value of $\big(\text{x}^2 +\text{y}^2 -\text{z}^2\big)^2 -\big(\text{x}^2 -\text{y}^2+\text{z}^2)^2$ is $4\text{x}^2(\text{y}^2-\text{z}^2).$
Using formula $(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x$,
we get $\left(x^2+y^2-z^2\right)^2-\left(x^2-y^2+z^2\right)^2$
$=(\text{x}^2)^2+(\text{y}^2)^2+(-\text{z}^2)^2+2(\text{x}^2)(\text{y}^2)+2(\text{y}^2)(-\text{z}^2)+2(-\text{z}^2)(\text{x}^2)\\-\Big[(\text{x}^2)^2+(-\text{y}^2)^2+(\text{z}^2)^2+2(\text{x}^2)(-\text{y}^2)+2(-\text{y}^2)(\text{z}^2)+2(\text{z}^2)(\text{x}^2)\Big]$
$=\text{x}^4+\text{y}^4+\text{z}^4+2\text{x}^2\text{y}^2-2\text{y}^2\text{z}^2-2\text{z}^2\text{x}^2\\-\big[\text{x}^4+\text{y}^4+\text{z}^4-2\text{x}^2\text{y}^2-2\text{y}^2\text{z}^2+2\text{z}^2\text{x}^2\big]$
By canceling the opposite terms,
we get$\big(\text{x}^2 +\text{y}^2 -\text{z}^2\big)^2 -\big(\text{x}^2 -\text{y}^2+\text{z}^2)^2=\not\text{x}^4+\not\text{y}^4+\not\text{z}^4\\+2\text{x}^2\text{y}^2-2\text{y}^2\text{z}^2-2\text{z}^2\text{x}^2\not\text{x}^4-\not\text{y}^4-\not\text{z}^4+2\text{x}^2\text{y}^2+2\text{y}^2\text{z}^2-2\text{z}^2\text{x}^2$
$=4\text{x}^2\text{y}^2-4\text{z}^2\text{x}^2$
Talking $4x^2$ as common factor we get$\big(\text{x}^2 +\text{y}^2 -\text{z}^2\big)^2 -\big(\text{x}^2 -\text{y}^2+\text{z}^2)^2=4\text{x}^2(\text{y}^2-\text{z}^2)$
Hence the value of $\big(\text{x}^2 +\text{y}^2 -\text{z}^2\big)^2 -\big(\text{x}^2 -\text{y}^2+\text{z}^2)^2$ is $4\text{x}^2(\text{y}^2-\text{z}^2).$
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