Given

$\rm 3^{x+3} + \frac{1}{3^{x}} - 28 = 0$

By using the law of indices, $\rm a^{m+n} = a^{m} \cdot a^{n}$,

$\rm or, 3^{x} \cdot 3^{3} + \frac{1}{3^{x}} - 28 = 0$

$\rm or, 3^{x} \cdot \left ( 3^{x} \cdot 3^{3} \right ) + 3^{x} \cdot \frac{1}{3^{x}} - 3^{x} \cdot 28 = 3^{x} \cdot 0$

$\rm or, 3^{x} \cdot 3^{x} \cdot 27 + 1 - 28 \cdot 3^{x} = 0$

By using the law of indices, $\rm a^{m+n} = a^{m} \cdot a^{n}$,

$\rm or, 3^{x + x} \cdot 27 + 1 - 28 \cdot 3^{x} = 0$

$\rm or,  27 \cdot 3^{2x} - 28 \cdot 3^{x} + 1 = 0$

By using the law of indices, $\rm  \left ( a^{m} \right )^{n} = a^{mn}$,

$\rm or, 27 \cdot \left ( 3^{x} \right) ^{2} - 28 \cdot 3^{x} + 1 = 0$

Let $\rm 3^{x} = a$. Substituting the supposed values in the above equation, we get,

$\rm or, 27 \cdot a^{2} - 28 \cdot a + 1 = 0$

The above equation is quadratic in a. We use the mid-term factorization method to solve the quadratic equation.

$\rm or, 27 \cdot a^{2} - (27 + 1) \cdot a + 1 = 0$

$\rm or, 27 \cdot a^{2} - 27 \cdot a - a + 1 = 0$

$\rm or, 27 a ( a - 1) - 1 (a - 1) = 0$

$\rm or, (27 a - 1) (a - 1) = 0$

Either

$\rm (27 a - 1) = 0$

$\rm or, 27a = 1$

$\rm or, a = \frac{1}{27}$

$\rm or, a = \frac{1}{3^{3}}$

$\rm or, a = 3^{-3}$

$\rm or, 3^{x} = 3^{-3}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = -3$

Or

$\rm (a - 1) = 0$

$\rm or, a = 1$

$\rm or, a = 3^{0}$

$\rm or, 3^{x} = 3^{0}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 0$

Hence, the required values of x are x = {-3, 0}.