Given the square of (2a-1) + [B-2] = 0, find the value of cube of a + cube of B
The square of ∵ (2a-1) + ┃ B-2 ┃ = 0
∴2a-1=0 b-2=0
a=1/2 b=2
∴a³+b³=(1/2)³+2³=1/8+8=65/8
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