The value of (xy−yx)(yz−zy)(zx−xz)(1x2−1y2)(1y2−1z2)(1z2−1x2) is
Consider the given expression.
⇒(xy−yx)(yz−zy)(zx−xz)(1x2−1y2)(1y2−1z2)(1z2−1x2)
⇒(x2−y2xy)(y2−z2yz)(z2−x2xz)(y2−x2x2y2)(z2−y2z2y2)(x2−z2x2z2)
⇒(x2−y2xy)(y2−z2yz)(z2−x2xz)−(x2−y2x2y2)(y2−z2z2y2)(z2−x2x2z2)
⇒−(x2y2xy)(z2y2yz)(x2z2xz)
⇒−x2y2z2
Hence, this is the answer.