a b 2 b c a c a b a b 2 a b c a b c , factorization

Lingqin is pure A B+C

Original formula 0 Therefore, the original formula contains the factor (b+c-a) because the original form is symmetrical.

The original form contains the equation (a+c-b), (b+a-c), so a 2 (b + c) + b 2 (a + c) + c 2 (a + b) -a 3-b 3-c 3-2abc k(a + b-c) (b + c -a) (c + a -b).

Let a=b=c=1

k 1 so.

a^2(b+c)+b^2(a+c)+c^2(a+b)-a^3-b^3-c^3-2abc＝－(a+b-c)(b+c-a)(c+a-b)

Order a b+c

Original 0 So the original contains the factor (b+c-a).

Since the original form is symmetrical.

The original style with a beat punch has a factored stool (a+c-b), (b+a-c), so. a^2(b+c)+b^2(a+c)+c^2(a+b)-a^3-b^3-c^3-2abc＝k(a+b-c)(b+c-a)(c+a-b)

Let a=b=c=1

k 1 so.

A 2(b+c)+b 2(a+c)+c 2(a+b)-a 3-b 3-c 3-2abc attack Zheng J (a+b-c)(b+c-a)(c+a-b).

A classic factorization provides an idea: take the last two formulas apart and make up (b-c) original formula = a 2(b-c) + (b 2)c-(c 2)b + a(c 2)-a(b 2).

a^2(b-c)+bc(b-c)-a(b+c)(b-c)=(b-c)[(a^2)-ab-ac+bc]=(b-c)(a-b)(a-c)

Method 1: Let a b x, a c y, b c z, then:

Original. ＝z（c＋x）（b＋y）－y（a＋z）（c－x）＋x（b－y）（a－z）

z（bc＋bx＋cy＋xy）－y（ac－ax＋cz－xz）＋x（ab－ay－bz＋yz）

bcz＋bxz＋cyz＋xyz－acy＋axy－cyz＋xyz＋abx－axy－bxz＋xyz

3xyz＋bcz－acy＋abx

3（a－b）（a－c）（b－c）＋bc（b－c）－ac（a－c）＋ab（a－b）

3（a－b）（a－c）（b－c）＋bc（b－c）－a^2c＋ac^2＋a^2b－ab^2

3（a－b）（a－c）（b－c）＋bc（b－c）＋a^2（b－c）－a（b^2－c^2）

3（a－b）（a－c）（b－c）＋bc（b－c）＋（b－c）［a^2－a（b＋c）］

3（a－b）（a－c）（b－c）＋（b－c）（bc＋a^2－ab－ac）

3（a－b）（a－c）（b－c）＋（b－c）［－c（a－b）＋a（a－b）］

3（a－b）（a－c）（b－c）＋（b－c）（a－b）（a－c）

4（a－b）（a－c）（b－c）。

Method 2: It is easy to verify that when a b, the original formula is 0;When a c, the original 0;When b c, the original formula is 0, and it is known from the remainder theorem that the original formula contains the equation (a b)(a c)(b c).

It is clear that the original is cubic and (a b)(a c)(b c) is cubic and the original k(a b)(a c)(b c) where k is a constant to be determined.

Let a 0, b 1, c 1, get:

k(a b)(a c)(b c) k(0 1)(0 1)(1 1) 2k,(b c)(a b c)(a b c) (1 1)(0 1 1)(0 1 1) 8,(c a)(b c a)(b c a) (1 0)(1 1 0)(1 1 0) 0,(a b)(c a b)( c a b) (0 1) ( 1 0 1) (1 0 1) 0, in this case 8, k 4.

So: 4(a b)(a c)(b c).