Simplified molecule

A 3+B 3+C 3-3 ABC uses the complete cubic formula to represent A and B.

= (a+b) 3-3a 2b-3ab 2+c 3-3abc uses the complete cubic formula again.

= (a+b+c) 3-3 (a+b) 2c-3 (a+b) c 2-3a 2b-3ab 2-3rd century BC

The latter items extract the common factor 3.

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

First, extract the common factor (a+b)c of the first two items in brackets.

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

Then extract the common factor ab of the last three terms.

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

Then the common factor (a+b+c) is extracted.

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

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

Then simplify the things in the right bracket, expand them all and omit a little, and get the result directly.

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

Now that the molecular simplification is completed, we can divide it.

Therefore, we can get a 2+b 2+c 2-ab-BC-AC = 3 from the known conditions.

Looking at the topic again, you need the value of (a-b) 2+(b-c) 2+(a-b) (b-c).

Simplify the formula and expand it all.

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

=a^2+b^2-2ab+b^2+c^2-2bc+ab-b^2-ac+bc

=a^2+b^2+C^2-ab-bc-ac

It can be seen that this formula is the same as the formula with simplified conditions above.

So (a-b) 2+(b-c) 2+(a-b) (b-c) = 3.

Have a rest. I'm tired of typing.

2. Let (a+b)/(a-b) = (b+c)/[2 (b-c)] = (c+a)/[3 (c-a)] = k.

Then a+b=k(a-b), b+c=2k(b-c), and c+a=3k(c-a).

Transform 8a+9b+5c as follows.

8a+9b+5c

=6(a+b)+3(b+c)+2(a+c)

Substitute the formula with k, and you can get

=6k(a-b)+6k(b-c)+6k(c-a)

=0

Complete the certificate.