What is x for x^3 - x^2 + 8x + 10 = 0 and x1 = 3i + 1?

Since the equation has a complex root, we'll recall the
property of complex roots: If a complex number is the root of an equation, then it's
conjugate is also the root of the equation.

So, we have as
root the number z = a + bi => z' = a - bi is also the root of the
equation.

Now, we'll verify if the complex numbers is the
root of the equation by substituting it into the original
equation.

We'll expand the cube using the
formula:

(a+b)^3 = a^3 + b^3 +
3ab(a+b)

a = 1 and b =
3i

(1+3i)^3 = 1^3 + (3i)^3 +
3*1*3i*(1+3i)

(1+3i)^3 = 1 - 27i +
9i(1+3i)

We'll remove the
brackets:

(1+3i)^3 = 1 - 27i + 9i -
27

We'll combine real parts and imaginary
parts:

(1+3i)^3 = -26 +
i(9-27)

(1+3i)^3 = -26 -
18i

We'll expand the square using the
formula:

(a+b)^2 = a^2 + 2ab +
b^2

(1+3i)^2 = 1^2 + 2*1*3i +
(3i)^2

(1+3i)^2 = 1 + 6i -
9

(1+3i)^2 = -8 + 6i

We'll
substitute the results into the expression (1+3i)^3 - (1+3i)^2 + 8(1+3i) + 10 =
0.

 -26 - 18i - (-8 + 6i) + 8 + 24i + 10 =
0

We'll combine like
terms:

(-26 + 8 + 8 + 10) + i(-18 - 6 + 24) =
0

0 + 0*i = 0

It
is obvious that 1 + 3i is the root of the
equation.

According to the rule, the
conjugate of 1 + 3i, namely 1 - 3i, is also the root of the equation. So it is not
necessary to verify if 1 - 3i is the root, since we've demonstrated that 1 + 3i is the
root of the equation.