(a - b)3 = (a - b)(a - b)(a - b)
(a - b)3 = a3 - 3a2b + 3ab2 - b3
You might note that the coefficients, 1, 3, 3, and 1 ( ignoring the signs for now)
are the third row of Pascals triangle. ( I call the first row the 0th row)
In case you forgot, here is how you construct the triangle, which I have written in left-justified form because it is easier to do that here.
To construct each entry in the next row, you look at the two entries above it (i.e. the one above it and to the right, and the one above it and to the left when you DON'T left justify it). At the beginning and the end of each row, when there's only one number above, put a 1.
1
11
121
1331
So the next row would be 1 and then 1+3=4, then 3+3=6, then 3+1=4 then 1.
This will tell you the coefficients of (a - b)3 and the next row is 1,5,10,10, 5,1 and this will tell you the coefficients of (a - b)4 All the coefficients of the terms are known as binomial coefficients. The binomial coefficients are how many terms there are of each kind. So now look at (a - b)3 as (a - b)(a - b)(a - b) as was done above.
We need aaa or a3 for the first term. We take 1 a from each term and there is only one such term so the coefficient is 1.
Now how about a2 b?
We need to take two a's from any of the three factors and then take the remaining b from whichever factor is unused. There are 3C2 or 3x2/2=3 ways to do that.
If you have not seen combinations or permutations just know that there are 3 choices for the first a and 2 choices for the second one. The multiplication principle tells us that there are 6 choices for the a's but it does not matter if which one we pick first, so we have to divide by 2 to take care of that. Bottom line, 3 choices for the a's and then the b is what is left.
Exact same thing with ab2
Now the last term is b3 and there is only 1 way to pick 3 b's from 3 factors. Of if you prefer
3C3=1
These would be the coefficient of (a+b)3 but of course we have a minus sign.
One easy way to take care of that is think of the term -b. When you have 1 of them it is still negative, when you have two you, have two negatives which is a positive and when you have 3 of them you have three negatives which is a negative.
Why go to all this trouble?
Because if you look at say (a-b)6 it is LOTS of work to multiply it out.
However, it is quite easy to compute or look up the terms in Pascals triangle and solve it in a few seconds. Chance of making an error this way is much less too.
a3 - b3 = (a - b) (a2 + ab + b2)
64 x cubed minus y cubed is (4x - y)(16x^2 + 4xy + y^2)
-1
It is simply subtraction of the whole number from one hundredth of the percent: that is, A% - B = A/100 - B or (100*A - B)/100. It may be possible to simplify the answer.
-116
a3 - b3 = (a - b) (a2 + ab + b2)
'a' minus 'b' whole cube is equal to 'a cube' minus 'b cube' minus (3 a square b ) plus (3 a b square) . .. .....thanks
X cubed - X cubed is zero.
64 x cubed minus y cubed is (4x - y)(16x^2 + 4xy + y^2)
(a-b)2 = a2 _ 2ab+b2
a2+b2+c2-2ab+2bc-2ca
-1
9 minus 8
(a+b)(a squared-ab+b squared)
It is simply subtraction of the whole number from one hundredth of the percent: that is, A% - B = A/100 - B or (100*A - B)/100. It may be possible to simplify the answer.
000000
-116