If $(4x - 5)^3 + (x - 2)^3 + 27(2x - 5)^3 = 9(4x -5)(x-2)(2x-5)$, then the value of $(x+\frac{3}{2})$ will be :

$\frac{7}{2}$

Given,

(4x - 5)³ + (x - 2)³+ 27(2x - 5)³= 9(4x - 5) (x - 2) (2x - 5)

We know,

If x + y + z = 0 then,

x³ + y³ + z³ = 3xyz

 (4x - 5)³ + (x - 2)³+ 27(2x - 5)³= 9(4x - 5) (x - 2) (2x - 5)

= (4x - 5)³ + (x - 2)³+ 27(2x - 5)³ - 3(4x - 5) (x - 2) 3(2x - 5) = 0

= (4x - 5) + (x - 2) + 3(2x - 5) = 0

= 4x - 5 + x - 2 + 6x - 15 = 0

= 11x = 22

= x = 2

Put the value of x in the required equation,

$(x+\frac{3}{2})$ = $(2+\frac{3}{2})$ = $\frac{7}{2}$