ok, here two:
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first one, easy way to calculate squares up to about 125^2 in tha head.
it works quite well if you still kind of remember the squares up to 25^2 from math-class. the formula is simple: x^2=(x+y)*(x-y)+y^2
the idea is to either make the (x+y) or the (x-y) something easy to multiply with.
examples: 11^2=12*10+1 (just as how-to

)
63^2=76*50+13^2=3800+169=3969
97^2=100*94+3^2=9400+9=9409
124^2=148*100+24^2=14800+576=15376
it's easiest if you get to multiply with 20, 50 and 100.
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other one: 3rd root (is that english? i mean x^(1/3)) of a 4-6 digit number (asumed it results a integer) in tha head.
things to remember:
1 x 1 x1=1
2 x 2 x 2=8
3 x 3 x 3=27
4 x 4 x 4=64
5 x 5 x 5=125
6 x 6 x 6=216
7 x 7 x 7=343
8 x 8 x 8=512
9 x 9 x 9=729
the idea is, that the 3rd root (i just call it that way) is a 2 digit-number.
first step: forget the last 3 digits. and chose the number from the list nearest under the left digits.
ex: 300 763^(1/3). forget last 3 digits->300 nearest one under 300 is 6^3=216, so the first digit of the result is a "6".
for the second digit of the result things to remember:
if the last digit of the given number (here 300763) ends on a 0, 1, 4 , 5 , 6 or 9, just take it as the result for the second digit.
if the last digit of the given number ends on 2, 3, 7 or 8, the result for the second digit is 10-(last digit).
in the example: 300763 end on "3" so the second digit of the result is (10-3)=7
so the third root of 300763=67
i hope you got it from my cranky english, but once you got the principles it's quite easy:
ex: 614125^(1/3)=
614->512->"8" as first digit of result.
614125 ends on 5= second digit of result
614125^(1/3)=85
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oh well... whatever... first one is quite a practical, second is something for nerd-partys...
(why do i still remember this sh147? it was like 10 years ago when i thought about such stuff (second one was on tv (no, not a US-channel and not BBC-learning

)...)
note to self: stop posting...