Given that 14a2 + 9b2 − a + 12b + 5 = 0, find the value of (a − 2) 2A2 − B2

Given that 14a2 + 9b2 − a + 12b + 5 = 0, find the value of (a − 2) 2A2 − B2

14a2+9b2−a+12b+5＝0，14a2-a+1+9b2+12b+4=0，（12a-1）2+（3b+2）2=0，∴12a-1=0，3b+2=0，∴a=2，b=-23，∴(a−2)2a2−b2=0．

Given 4A & # 178; + 9b & # 178; - 4A + 12b + 5 = 0, find the value of ab

(4a^2 - 4a + 1) + (9b^2 + 12b + 4) = 0
(2a - 1)^2 + (3b + 2)^2 = 0
To make the above formula true, we must:
2a - 1 = 0,3b + 2 = 0
That is: a = 1 / 2, B = - 2 / 3
Therefore, ab = - 1 / 3

Decomposition factor: - x ^ 3 + x = ② a ^ 2-4a + 3

-x^3+x=x(1-x^2)=x(1+x)(1-x)
a^2-4a+3=(a-1)(a-3)

It is known that a = {x | x ^ 2 + 4ax-4a + 3 = 0}, B = {x} x ^ 2 + (A-1) x + A ^ 2 = 0}, C = {X
Given a = {x | x ^ 2 + 4ax-4a + 3 = 0}, B = {x | x ^ 2 + (A-1) x + A ^ 2 = 0}, C = {x | x ^ 2 + 2ax-2a = 0}, at least one set is not an empty set, the value range of real number a is obtained
Because I am rather slow, so I must be detailed. I'd better not jump,

A：△＝16a²+16a－12＝4（2a－1）（2a+3）≥0,
A ≥ 1 / 2 or a ≤ - 3 / 2
B：△＝（a－1）²－4a²＝－（1+a）（3a-1）≥0
∴-1≤a≤1/3
C：△＝4a²+8a＝4a（a+2）≥0
A ≥ 0 or a ≤ - 2
When a ∈ R, at least one set is empty