#### Answer

(a) $f(g(x))=-64x^3+48x^2-12x-1$; Domain : $(-\infty, \infty).$
(b) $g(f(x))= 9-4x^3$; Domain : $(-\infty, \infty).$
(c) $f(f(x))= x^9-6x^6+12x^3-10$; Domain : $(-\infty, \infty).$
(d) $g(g(x))= 16x-3$; Domain : $(-\infty, \infty).$

#### Work Step by Step

$f(x)=x^3-2$
$g(x) = 1-4x$
(a) $f(g(x))= (1-4x)^3-2 = 1-12x+48x^2-64x^3-2= -64x^3+48x^2-12x-1$
As we have no restrictions, the domain is $(-\infty, \infty).$
(b) $g(f(x))= 1-4(x^3-2) = 1-4x^3+8 = 9-4x^3$
As we have no restrictions, the domain is $(-\infty, \infty).$
(c) $f(f(x))= (x^3-2)^3-2= x^9-6x^5+12x^3-8-2 = x^9-6x^6+12x^3-10$
As we have no restrictions, the domain is $(-\infty, \infty).$
(d) $g(g(x))= 1-4(1-4x) = 1-4+16x=16x-3$
As we have no restrictions, the domain is $(-\infty, \infty).$