Switching the base

$Given\quad that\\ \qquad \qquad { 3 }^{ x }\quad =\quad { 81 }\\ or\qquad \quad { 3 }^{ x }\quad =\quad { 3 }^{ 4 }\\ It\quad shows\quad that\\ \qquad \qquad \boxed { x=4 } \\ Now\\ \qquad \qquad { x }^{ 3 }\quad =\quad 4^{ 3 }\qquad \because \quad \boxed { x=4 } \\ Thus\quad answer\quad is\quad \boxed { 64 }$

$\large 3^x = 81 \\ \log 3^x = \log 3^4 \\ x\log 3 = 4 \log 3 \\ x = 4 \\ \therefore x^3 = 4^3 =\huge\boxed{64}$

Since $3^x = 81$ and $81 = 9^2 = (3^2)^2 = 3^4$

$3^x = 3^4$

$x = 4$

$4 ^ 3 = 4 \cdot 4 \cdot 4 = 16 \cdot 4 = 64$

The answer is $64$

3 to the power of 4 is 81, so x=4 Then substitute 4 into the equation x^3.

When 4^3, we get 64.

Easy as pie, INTELLIGENT minds not necessary here.

3^x=81,

But, we know

3^4=81

So,

x=4

Now,

4^3=64

3^x = 81 = 3^4 x = 4 So, 4^3 = 64