The given equation is a cubic equation of the form ax³ + bx² + cx + d = 0, where a = 1, b = -7, c = -21, and d = -27. We need to find the roots of this equation.



The Rational Root Theorem states that if a polynomial with integer coefficients has a rational root p/q (in lowest terms), then p must divide the constant term and q must divide the leading coefficient.

In this case, the constant term is -27 and the leading coefficient is 1. Therefore, the possible rational roots are:

±1, ±3, ±9, ±27

We can use synthetic division to check which of these possible roots are actual roots of the equation.



Using synthetic division with the possible root -3, we get:

-3 | 1 -7 -21 -27
| -3 30 -27
|-------------
1 -10 9 -54

Since the remainder is not 0, -3 is not a root of the equation.

Using synthetic division with the possible root -1, we get:

-1 | 1 -7 -21 -27
| -1 8 13
|-------------
1 -8 -13 -14

Since the remainder is not 0, -1 is not a root of the equation.

Using synthetic division with the possible root 3, we get:

3 | 1 -7 -21 -27
| 3 -12 -9
|-------------
1 -4 -33 -36

Since the remainder is not 0, 3 is not a root of the equation.

Using synthetic division with the possible root 9, we get:

9 | 1 -7 -21 -27
| 9 18 -27
|-------------
1 2 -3 -54

Since the remainder is not 0, 9 is not a root of the equation.

Using synthetic division with the possible root -9, we get:

-9 | 1 -7 -21 -27
| -9 144 -99
|--------------
1 -16 123 72

Since the remainder is not 0, -9 is not a root of the equation.

Using synthetic division with the possible root -27, we get:

-27 | 1 -7 -21 -27
| 27 540 -585
|--------------
1 20 519 -612

Since the remainder is not 0, -27 is not a root of the equation.

Therefore, none of the possible rational roots are actual roots of the equation.



Since the equation is a cubic equation, it must have 3 roots (including complex roots). Since none of the possible rational roots are actual roots, the remaining roots must be complex.

Using the cubic formula, we