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Just draw one line for each digit place and write the number next to it. The cross section where the two lines meet will be this times that. I write that number down on that dot respectfully.
Right to Left: Carry the 8 to 162 to make it 170. Carry the 7 to 243 to make 250. Save the 1. Carry the 5 to 162 to make 167. Save the 2. Carry the 6 to 81 to make 87. Save the 1. You get 877001 with the left over digits. Add 121 (saved digits) to the high-end digit placing, the 877. You get 998001.
I draw one line for each digit and put the number next to it. The number of dots at the cross section would be this times that... so I write that number next to the dot.
This method is mostly only useful for those who don't know their basic times tables, and is extremely tedious when performing problems with larger numbers. (ie. 999 x 999) While it is still possible using this method it requires about half a page of paper if used efficiently.
To all of those complaining about 0's use dotted lines, or simply another type of line which you can distinguish as 0, and then anytime you come across a dotted line, know that the number will be 0, because 0 x # is equal to 0.
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The numbers you get from left to right are:
81,162, 243, 162, 81
Right to Left:
Carry the 8 to 162 to make it 170.
Carry the 7 to 243 to make 250. Save the 1.
Carry the 5 to 162 to make 167. Save the 2.
Carry the 6 to 81 to make 87. Save the 1.
You get 877001 with the left over digits.
Add 121 (saved digits) to the high-end digit placing, the 877.
You get 998001.