How do you derive the equation a b 3, a b 4, a b 5 in mathematics?

This is very clear in the high school binomial theorem, the coefficients are all combinations, and the result of the addition of the coefficients is also the corresponding power of 2.

The coefficient to the power of 3 is c30=1 c31=3 1=3 c32=3*2 2*1=3 c33=3*2*1 3*2*1=1

o(∩_o~

Derived from the binomial theorem.

a+b)^2=a^2+2ab+b^2

a+b) 3=a 3+3a 2b+3ab 2+b 3 rule: the coefficient of the latter terms.

And so on and so forth.

Yang Hui Triangle: where The first row represents the coefficient of each term of the zero mode 1 of (a+b).

The second line represents the coefficient of each term of the primary method a+b of (a+b).

The third line represents the quadratic way (a+b) a 2 + 2 ab + b 2 coefficients for each term.

And so on. So the formula of the cubic (a+b) is.

a 3+3a 2b+3ab 2+b 3 (fourth row) if it is a cubic of (a-b), it is: a 3-3a 2b+3ab 2-b 3 (that is, the plus sign in front of the term containing b to the odd power becomes a minus sign).

Note: The number after the "," is the exponent before the letter of "".

a+b)^3=a^3+3*a^2*b+3*a*b^2+b^3a-b)^3=a^3-3*a^2*b+3*a*b^2-b^3a+b)^3=(a+b)*(a+b)*(a+b)[(a+b)*a+(a+b)*b]*(a+b)(a^2+b^2+2ab)*(a+b)

a^2+b^2+2ab)*a+(a^2+b^2+2ab)*ba^3+b^3+3ab^2+3a^2b

a+b)^3=a^3+b^3+3ab(a+b)

Yang Hui Triangle:

Thereinto. The first row represents the coefficient for each term of the zero method 1 of (a+b).

The second line represents the coefficient of each term of the primary method a+b of (a+b).

The third line represents the quadratic way (a+b) a 2 + 2 ab + b 2 coefficients for each term.

And so on. So the formula of the cubic (a+b) is.

a 3+3a 2b+3ab 2+b 3 (fourth row) if it is the third power of (a-b), it is a 3-3a 2b+3ab 2-b 3 (that is, the maga sign in front of the term containing b to the odd power becomes a minus sign).

Note: The number after the "" is the exponent of the first letter of the "dusty reed".

a+b) x of the public code letter bright type.

Binomial theorem.

Delay width a+b) x

c(x,0)a^xb^0+c(x,1)a^(x-1)b^1+..c(x,x)a^0b^x

Note: c(x,0) tanwang = 1

c(x,1)=x

I calculated it by hand, and there was no plagiarism.

a+b）^n

a+b)^3=a^3+3a^2b+3ab^2+b^3

a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4

a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5

a+b)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6

The following is omitted, I calculated that n=10 is too difficult to hit, so I won't play.

Index (n=) Number of letter exponents.

2n-1 (2n-1) (2n-2 1) …1 2n-2) (2n-1) 2n

2n 2n (2n-1 1)……2n-1 1) 2n 2n+1

The following are the indices, some of them, not all of them.

Indices Coefficients.

2n-1 1 2n-1 x1 y1 z1 a1

2n 1 2n x2 y2 z2 a2

2n-1)：x1=1：(n-1) x1：y1=3：(2n-3) y1：z1=4：(2n-4) z1：a1=5：(2n-5)

2n：x2=2：（2n-1） x2：y2=3：(2n-2) y2：z2=4：(2n-3) z2：a2=5：(2n-4)

The latter ones haven't been counted yet, only these.

a+b) x.

Binomial theorem.

a+b）^x

c(x,0)a^xb^0+c(x,1)a^(x-1)b^1+..c(x,x)a^0b^x

Note: c(x,0)=1

c(x,1)=x

(a-b)³=a³-3a²b+3ab²-b³

a+b)³=a³+3a²b+3ab²+b³

It can be made using the binomial theorem.

The original question should be a 3-b 3=(a-b)(a 2+ab+b 2), and the derivation method is to get the left from the right.

Similarly, there are two formulas: a 3+b 3=(a+b)(a 2-ab+b 2), but the symbols are different in some places.

Backward: (a (m) * b (0) + a (m-1) * b (1) +a (0)*b (m)) a-b) (multiply a and b by the left equation respectively).

Written in two lines, you can see the relationship)

a^(m+1)*b^(0) +a^(m)*b^(1) +a^(m-1)*b^(2) +a^(1)*b^(m)

a^(m)*b^(1) -a^(m-1)*b^(2) -a^(1)*b^(m) -a^(0)*b^(m+1)

a^(m+1)*b^(0) -a^(0)*b^(m+1)=a^(m+1) -b^(m+1)