（1）6（m-n）³-12（n-m）² （2）3a（a-2b）²-18b（2b-a）²

（1）6（m-n）³-12（n-m）² （2）3a（a-2b）²-18b（2b-a）²

one
The original formula = 6 (m-n) &# 179; - 12 (m-n) &# 178;
=6（m-n）²(m-n-2)
two
The original formula is 3A (a-2b) & # 178; - 18B (a-2b) & # 178;
=3（a-2b）²(a-6b)

In the following figure, the bottom of the cylinder is divided into several equal sectors and cut into an approximate cuboid. 1. The sum of the areas of the front and back sides of the cuboid is the area of the cylinder. 2. The cuboid has a width of 2 cm and a height of 5 cm. The volume of the cylinder is a cubic centimeter

The sum of the areas of the front and back sides of a cuboid is the [side area] of a cylinder
The cuboid is 2 cm wide and 5 cm high, and the volume of the cylinder is [62.8] cubic cm
3.14 * 2 * 2 * 5 = 62.8 cm3

Please come to solve this problem for me
When a arrives at B, B is 30 kilometers away from A. when B arrives at a, a is 40 kilometers away from B. how many meters is the distance between a and B?

That is, in the 30 km time of B line, the 40 km time of a line
So the speed ratio is 40:30 = 4:3
So the whole process of a line
B / 3 / 4 OK‘
30÷(1-3/4)=120
A: the AB distance is 120 km