Given equation
(64/125)a³ – 8 – (96/25)a² + (48/5)a
We will simply and write in the cube format
= (4/5)³a³ – (2)³ – 6(4/5)²a² + 12(4/5)a
=(4a/5)³ – (2)³ – 6*(4a/5)² + 12(4a/5)—————(1)
We will assume 4a/5 = x
Substituting the assumed value of x in equation (1)
= x³ – (2)³ -6x² + 12x
= x³ – -6x² + 12x – (2)³ ————–(2)
The above equation is the form of the identity (a-b)³
(a-b)³ = a³ – 3a²b + 3ab² – b³
Expressing the equation (2) as per the identity, we get
= x³ – (3*x²*2) + (3*x*2²) – (2)³ ———(3)
The equation 3 is the form of [(a-b)³ = a³ – 3ab(a-b) – b³]
Simplyfing as per the above identity, we get
= x³ – 6x(x-2) – (2)³
= (x-2)³
Putting x = 4a/5 we get
= [(4a/5) – 2]³
Taking out 2 as a common factor
= (2³){(2a/5) – 1}³
=8{(2a – 5)/5}³
= 8/5³ (2a – 5)³
= (8/125) (2a – 5) (2a – 5) (2a – 5)
= (8/125) (2a – 5)³
