We are asked to solve the equation $2^{3n+2} - 7 \times 2^{2n+2} - 31 \times 2^n - 8 = 0$ for $n \in R$.

1. Problem Description

We are asked to solve the equation $2^{3n+2} - 7 \times 2^{2n+2} - 31 \times 2^n - 8 = 0$ for $n \in R$.

2. Solution Steps

First, rewrite the equation using exponent rules:

$2^{3n+2} = 2^{3n} \times 2^2 = 4 \times (2^n)^3$

$2^{2n+2} = 2^{2n} \times 2^2 = 4 \times (2^n)^2$

Substituting these into the original equation gives:

$4 \times (2^n)^3 - 7 \times 4 \times (2^n)^2 - 31 \times 2^n - 8 = 0$

$4 \times (2^n)^3 - 28 \times (2^n)^2 - 31 \times 2^n - 8 = 0$

Let $x = 2^n$. Then the equation becomes:

$4x^3 - 28x^2 - 31x - 8 = 0$

We can try integer roots of the form $\pm \frac{factors \; of \; 8}{factors \; of \; 4}$.

Possible rational roots are $\pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{2}, \pm \frac{1}{4}$.

Trying $x = 8$:

$4(8^3) - 28(8^2) - 31(8) - 8 = 4(512) - 28(64) - 248 - 8 = 2048 - 1792 - 248 - 8 = 2048 - 2048 = 0$

So, $x = 8$ is a root. Thus, $(x-8)$ is a factor.

We can perform polynomial division to find the other factor:

$(4x^3 - 28x^2 - 31x - 8) / (x - 8) = 4x^2 + 4x + 1$

Thus, the equation becomes:

$(x - 8)(4x^2 + 4x + 1) = 0$

$(x - 8)(2x + 1)^2 = 0$

So, $x = 8$ or $2x + 1 = 0 \implies x = -\frac{1}{2}$.

Since $x = 2^n$, we have $2^n = 8$ or $2^n = -\frac{1}{2}$.

$2^n = 8 = 2^3$, so $n = 3$.

$2^n = -\frac{1}{2}$ has no real solution since $2^n$ is always positive.

Therefore, the only real solution is $n = 3$.

3. Final Answer

$n = 3$