Just a simple cubic

Since $a$ is a real number, then $P(x)$ is a real polynomial, and thus $-i-3$ is another root of $P(x)$ . Now, since $(x-(i-3))$ and $(x-(-i-3))$ are two linear factors of $P(x)$ , we can write:

$P(x)=a\big(x-(i-3)\big)\big(x-(-i-3)\big)(x-\beta)=a(x^2+6x+10)(x-\beta)$ Where $\beta$ is the real root.

Now, factorizing the main coefficient in the original expression, we have: $P(x)=a\bigg(x^3+\frac{9}{a}x^2+x-\frac{30}{a}\bigg)$

From equating both expressions for $P(x)$ , we get:

$a\bigg(x^3+\frac{9}{a}x^2+x-\frac{30}{a}\bigg)=a\bigg(x^3+(6-\beta)x^2+(10-6\beta)x-10\beta\bigg)$ Equating coefficients of $x^2$ and independent terms, we get:

$-\frac{30}{a}=-10\beta \Rightarrow a\beta=3$ $\frac{9}{a}=6-\beta \Rightarrow a\beta=6a-9$

Solving simultaneously for $a$ and $\beta$ , we get $a=2$ and $\beta=\frac{3}{2}$ . Hence:

$a+\beta=2+\frac{3}{2}=\boxed{\frac{7}{2}}$

Since P(x) as described is of all real coefficients,

x = -3 + $\sqrt{-1}$

$\implies (x + 3)^2 = -1$

$\implies x^2 + 6 x + 10 = 0$

Since $\alpha \gamma = 10,$ {Here is the trap of this question for -10 by mistake of other way.}

$\frac{30}{a} = \alpha \beta \gamma = 10 \beta.$

Hence $\beta = \frac{3}{a}.$

(x - $\frac{3}{a}$ )( $x^2 + 6 x + 10$ ) = $0$

$\implies a x^3 + (6 a - 3) x^2 + (10 a - 18) x - 30 = 0$

6 a - 3 = 9 $\implies$ a = 2

Checked that 10 a - 18 = 10 (2) - 18 = 2 = a which is correct.

$\beta = \frac{3}{a} = \frac32$

$a + \beta = 2 + \frac32 = \frac72$

Answer: $\boxed{\frac72}$