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Using the formula to calculate (2a + 1) ^ 2 - (1-2a) ^ 2
（2a+1）^2-(1-2a)^2
=[(2a+1)+(1-2a)][(2a+1)-(1-2a)]
=2×4a
=8a
Given the equation (a + 2) C, C = 1 does not hold, find the square of a - 2A + 1
the sooner the better,
Equation (a + 2) C = a + 2
C is not equal to 1,
So a + 2 = 0
A=-2
The square of a - 2A + 1
=(A-1)^2
=(-3)^2
=9
If B = 1 does not hold from the equation (a + 3) B = a + 3, find the square of 2a-2a-2
From the equation (a + 3) B = a + 3, B = 1 does not hold
It shows that both sides of the equation cannot be divided by a + 3 at the same time
a+3=0
a=-3
2a^2-2a-2
=2(a^2-a-1)
=2*(9+3-1)
=2*11
=22
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