Known a b c 5 to find the value of a a b c b c a b c b c a detailed process

Simplification of a(a-b-c)+b(c-a+b)+c(b+c-a)a(a-b-c) to a a-(b+c) b(c-a+b) to -b a-(b+c) c(b+c-a) to -a(a-b-c) to -a(a-b-c) to -a(a-b-c) to -a(a-b-c) to a.

a a-(b+c) -b a-(b+c) -c a-(b+c) a-b-c=-5.

a-b-c）*（5）

Then bring in (-5) * (5) to get 25

You can first assign values to a and b to find c

For example. Let a=2 and b=1 be substituted into the original formula, and c can be obtained

Then substitute a, b, and c into the required algebraic formula.

a+b) c=(a+c) b=(b+c) a=kLet the above equation be equal to k, get.

a+b=kc

a+c=kb

b+c=ka

The above three types of beam rent are added together.

2(a+b+c)=k(a+b+c)

k(a+b+c)-2(a+b+c)=0

k-2)(a+b+c)=0

The solution is: k=2 and a+b+c=0, when k=2, (a+b)(a+c)(b+c) abc=2*2*2=8;

When a+b+c=0, the cavity spike can be obtained: a+b=-c, a+c=-b, b+c=-a, then.

a+b)(a+c)(b+c)/abc=-c*(-b)*(a)/abc=-1.

Because. It is known that the branch is a+b c=b+c a=a+c b, and the travel delay a+b+c≠0, so.

a+b/c=b+c/a=a+c/b=(a+b+b+c+a+c)/(a+b+c)=2

i.e. a+b=2c,b+c=2a,a+c=2ba+c)(a+b)(b+c) violently destroy abc2c 2a 2b abc

Let a+b=2x (1).

b+c=2y (2)

c+a=2z (3)

Then: a+b+c=x+y+z (4).

1)-(2):a-c=2(x-y) (5)2)-(3):b-a=2(y-z) (6)3)-(1):c-b=2(z-x) (7)4)-(2):a=x+z-y (8)

4)-(3):b=y+x-z (9)

4)-(1):c=y+z-x (10)

Then the line is in the volt: file carrying (a-b) (b-c) (c-a) (a+b) (b+c) (c+a).

2(z-y)2(x-z)2(y-x)/2x2y2z(zxy-zxx-zzy+zzx-yxy+yxx+yzy-yzx)/xyz

x Peihe y-z x+z y-y z+x z+y x=5 132 then: a (a+b)+b (b+c)+c (c+a)(x+z-y) x+(y+x-z) y+(y+z-x) z1+z x-y x+1+x y-z y+y z+1-x z

Set the side (a+b-c) c=(a-b+c) b=(-a+b+c) a=k, then there is a mausoleum:

a+b-c=kc

a-b+c=kb

a+b+c=ka

The above three types of phase socks are added virtually, and you get:

a+b+c=k(a+b+c)

k(a+b+c)-(a+b+c)=0

k-1)(a+b+c)=0

Available: k=1, then there is.

a+b-c=c, get: a+b=2c, a-b+c=b, get: c+a=2b, a+b+c=a, get: b+c=2a, so:

a+b)(b+c)(c+a)/abc

2c*2a*2b/abc

8abc/abc

a+b+c=0, there is.

a+b=-c

b+c=-a

c+a=-b

So. a+b)(b+c)(c+a)/abc-c*(-a)(-b)/abc

abc/abc

In summary, there are two values of (a+b)(b+c)(c+a) abc, which are: 8 and -1

If a+b+c=0, then a+b=-c, a+c=-b, b+c=-a, original = (-c)(-a)(-b) abc=-1

If a+b+c≠0, it is obtained from the proportional property: a b+c=b c+a=c a+b=(a+b+c) (2a+2b+2c)=1 2

So a+b=2c, b+c=2a, a+c=2b, original = (2c)(2a)(2b) abc=8

So the original value is -1 or 8

Let a b+c=b c+a=c a+b=t

then we get a=(b+c)t

b=(a+c)t

c=(b+a)t

The addition of three formulas (a+b+c)=2(a+b+c)*t is divided into two cases:

1) When a+b+c =0 and t is any real number, then a+b=-c, b+c=-a, c+a=-b(a+b)(b+c)(c+a) abc=(-c)*(a)*(b) (abc).

-abc)/(abc)

1(2) when a+b+c ≠ 0 (a+b+c)=2(a+b+c)*t both sides are divided by (a+b+c).

t=1 2 then a b+c=b c+a=c a+b=1 2 dissolves to obtain b+c=2a, c+a=2b, a+b=2c(a+b)(b+c)(c+a) abc=(2c)*(2a)*(2b) (abc).

8abc)/(abc)

8'Sum up.

a+b)(b+c)(c+a) abc has a value of -1 or 8

Let b+c a=c+a b+a=c b+a=t, then b+c=at c+a=bt a+bt a+b=ct +=2(a+b+c)=(a+b+c)t(1) when a+b+c is not 0.

t=2 b+c=2a c+a=2b a+b=2cc=2a·2b·2cABC=8

2) When a+b+c=0.

a+b=-c

Original = -1

Solution: Let (b+c) a=(c+a) b=(a+b) c=k b+c=ak,c+a=bk,a+b=ck 2(a+b+c)=k(a+b+c).

If a+b+c≠0, then k=2

If a+b+c=0, then a+b=-c k=(a+b) c=-c c=-1

Primitive = abc (kc*ka*kb) = 1 k When k = 2, primitive = 1 8;When k = -1, the original formula = -1

If a+b+c=0, then a+b=-c, a+c=-b, b+c=-a, original =abc (a+b)(b+c)(c+a)=-1

If a+b+c≠0, it is obtained from the proportional property: a b+c=b c+a=c a+b=(a+b+c) (2a+2b+2c)=1 2

So a+b=2c,b+c=2a,a+c=2b,original formula=abc (2a)(2b)(2c)=1 8

So the original value is -1 or 1 8

Hello, Shao Wenchao of the potato group will answer your questions and solve problems, if you don't understand anything about this question, you can ask, if you are satisfied, remember to adopt. It is not easy to answer the question, please understand, thank you. And good luck with your learning!