Ask math questions a + 2B + 3C = 12, AA + BB + CC = AB + BC + AC, find the value of AA + BB + CC

Ask math questions a + 2B + 3C = 12, AA + BB + CC = AB + BC + AC, find the value of AA + BB + CC

aa+bb+cc-ab-bc-ac=0
The combination of the above formula × 2 yields
(aa-2ab+bb)+(aa-2ac+cc)+(bb-2bc+cc)=0
The square formula is obtained
a=b=c
a+2b+3c=6a=12
a=2
aa+bb+cc=12

If: A.B, C are the three sides of △ ABC, try to judge whether the value of (AA + BB CC) (AA + BB CC) - 4aabb is positive or negative

The square difference formula = (A & sup2; + B & sup2; - C & sup2; + 2Ab) (A & sup2; + B & sup2; - C & sup2; - 2Ab) = [(a + b) & sup2; - C & sup2;] [(a-b) & sup2; - C & sup2;] = (a + B + C) (a + B-C) (a-b + C) (a-b-c) the side length is greater than 0A + B + C > 0, the sum of the two sides of the triangle is greater than the third side, so a + B-C > 0a-b + C > 0A

Given a + B + C = 0, ABC = - 15, find AA (B + C) + BB (a + C) + CC (a + b)

If a + B + C = 0, then a + B = - C; B + C = - A; C + a = - B
a²(b+c)+b²(a+c)+c²(a+b)
=a²b+a²c+ab²+b²c+ac²+bc²
=(a²b+ab²)+(a²c+ac²)+(b²c+bc²)
=ab(a+b)+ac(a+c)+bc(b+c)
=ab×(-c)+ac×(-b)+bc×(-a)
=-3abc
=(-3)×(-15)
=45