The following is a detailed answer
Adopt in detail
The multiplication formula 1. Multiplication formula is also called simple multiplication formula, which sums the multiplication results of some special polynomials directly.
Apply. Each letter in the formula can generally represent numbers, monomials, polynomials, and some can be extended to fractions.
Radical form
Formulas can be used not only from left to right (multiplication expansion), but also from right to left (factorization).
Remember some important formula variants and their inverse operations-division and so on.
2. The basic formula is the most commonly used and basic formula, from which other formulas can be derived.
Complete square formula: (A B) 2 = A2AB+B2
Square difference formula: (a+b)(a-b)=a2-b2,
Cubic sum (difference) formula: (a b) (a2mab+B2) = a3 B3.
3. Generalization of the formula:
① Polynomial square formula: (a+b+c+d) 2 = a2+b2+c2+d2+2ab+2ac+2ad+2bc+2bd+2cd.
That is, the square of the polynomial is equal to the sum of the squares of the terms, plus twice the product of every two terms.
② binomial theorem: (ab) 3 = A332B+3B2B3
(a b)4=a4 4a3b+6a2b2 4ab3+b4，
(a b)5 = a55a4b+ 10a3b 2 10a2b 3+5ab 4 b5，
…………
Pay attention to the laws of terms, exponents, coefficients and symbols of right expansion.
③ Formulas derived from square difference, cubic sum (difference) formulas.
(a+b)(a3-a2b+ab2-b3)=a4-b4，
(a+b)(a4-a3b+a2b2-ab3+b4)=a5+b5，
(a+b)(a5-a4b+a3 B2-a2 B3+ab4-b5)= a6-B6，
…………
Pay attention to the law of the number, index, coefficient and symbol of the second factor on the left.
Under the condition of positive integer exponent, it can be summarized as follows: let n be a positive integer.
⑴(a+b)(a2n- 1-a2n-2 b+a2n-3 B2-…+ab2n-2-b2n- 1)= a2n-b2n，
⑵(a+b)(a2n-a2n- 1b+a2n-2 B2-…-ab2n- 1+b2n)= a2n+ 1+b2n+ 1，
Similarly:
⑶(a-b)(an- 1+an-2 b+an-3 B2+…+ABN-2+bn- 1)= an-bn。
4. Deformation of formula and its inverse operation
a2+B2 =(a+b)2 = a2+2ab+B2； (a-b)2=(a+b)2-4ab。
a3+B3 =(a+b)3 = a3+3a2b+3ab 2+B3 = a3+B3+3ab(a+b)-3ab(a+b)。
From the generalization of the formula, we can know that when n is a positive integer, an-bn can be divisible by a-b;
A2n+ 1+b2n+ 1 is divisible by a+b; A2n-b2n is divisible by a+b and A-B. 。
Example b
Example 1. Known: X+Y = A, xy = b.
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Found:1x2+y2; ②x3+y3； ③x4+y4； ④x5+y5。
Solution: ① X2+Y2 = (x+y) 2-2xy = A2-2b;
②x3+y3 =(x+y)3-3xy(x+y)= a3-3ab；
③x4+y4 =(x+y)4-4xy(x2+y2)-6x2y 2 = a4-4a2b+2 B2；
④X5+y5 =(x+y)(x4-x3y+x2 y2-xy3+y4)
=(x+y)[x4+y4-xy(x2+y2)+x2y2]
=a[a4-4a2b+2b2-b(a2-2b)+b2]
=a5-5a3b+5ab2。
Example 2. It is proved that the sum of the product of four consecutive integers plus 1 must be the square of integers.
Prove: Let these four numbers be a, a+ 1, a+2 and A+3(A is an integer).
a(a+ 1)(a+2)(a+3)+ 1 = a(a+3)(a+ 1)(a+2)+ 1
=(a2+3a)(a2+3a+2)+ 1
=(a2+3a)2+2(a2+3a)+ 1
=(a2+3a+ 1)2。
∵a is an integer, and the sum, difference, product and power of integers are also integers.
A2+3a+ 1 is an integer.
Example 3. It is proved that 2222+311is divisible by 7.
Proof: 2222+311= (22)11+311= 465438.
∫a2n+ 1+b2n+ 1 is divisible by a+b (see executive summary 4).
∴ 4111+311is divisible by 4+3.
∴ 2222+311is divisible by 7.
Example 4. The calculation rule of "square number of two digits with unit number of 5" is derived by using the complete square formula.
Solution: ∫ (10a+5) 2 =100a2+2×10a× 5+25 =100a (a+1)+25.
∴ "Square number of two digits and five digits" is characterized by:
The last two digits of the power are the square of the single digit 5 of the base, and the digits above the hundred digits of the power are the digits above the ten digits of the base.
The word a is multiplied by the product of (a+ 1).
For example: 152=225, and the number on the hundredth power is 2 =1× 2;
252=625, 6=2×3;
352= 1225, 12=3×4;
……
1052= 1 1025, 1 10= 10× 1 1.