Happy Holiday

It is given that $f(x)=x^5+9x^4+7x^3+16x^2+6x+7$ . By trial and error, we have:

\(\begin{array} {} f(\sqrt{-1}) & = f(i) & =i+9-7i-16+6i+7 & = 0 \\ f(-\sqrt{-1}) & = f(-i) & =-i+9+7i-16-6i+7 & = 0 \end{array} \)

$\Rightarrow (x-i)(x+i)=x^2+1$ is a factor of $f(x)$ and $f(x)=(x^2+1)(x^3+9x^2+6x+7)$ .

Therefore the real root must come from $x^3+9x^2+6x+7=0$ , which we can solved for using Cardano Formula as follows.

$x^3+9x^2+6x+7=0 \quad \Rightarrow (x+3)^3-21(x+3)+43=0$

Let $x+3=u^3+v^3$ and $3uv-21=0$ then

$\begin{aligned} x^3+9x^2+6x+7 & =u^3+v^3+3uv(u+v) -21(u+v)+43 \\ & =u^3+v^3+(3uv-21)(u+v) +43 \\ & = u^3+v^3 + 43 = 0 \end{aligned}$

$\Rightarrow \begin{cases} u^3+v^3 =- 43 \\ u^3v^3 = 7^3= 343 \end{cases} \Rightarrow u^3$ and $v^3$ are roots of $z^2+43z+343=0$

$\Rightarrow \begin{cases} u^3 = \dfrac {-43+\sqrt{43^2-4(343)}}{2} = -\dfrac {43-3\sqrt{53}}{2} \\ v^3 = -\dfrac {43+3\sqrt{53}}{2} = -\dfrac {43^2-9(53)}{2(43-3\sqrt{53})} = -\dfrac {686}{43-3\sqrt{53}} \end{cases}$

$\Rightarrow \begin{cases} u = - \sqrt [3] {\dfrac {43-3\sqrt{53}}{2}} \\ v = -7\sqrt [3] {\dfrac {2}{43-3\sqrt{53}}} \end{cases}$

$\Rightarrow x = -3+v+u = -3-7\sqrt [3] {\dfrac {2}{43-3\sqrt{53}}} - \sqrt [3] {\dfrac {43-3\sqrt{53}}{2}}$

$\Rightarrow a+b+c+1=3+43+53+1 = \boxed{100}$

I will present you 2 methods, first is the long one and second is the short one.

$\large \text{Method-1}$

Watch a little see a little it gets factorised neatly as follows :

$f(x) = x^{5}+9x^{4}+7x^{3}+16x^{2}+6x+7=(x^{5}+7x^{3}+6x)+(9x^{4}+16x^{2}+7)$

$\Rightarrow f(x) = x(x^{2}+1)(x^{2}+6)+(9x^{2}+7)(x^{2}+1)$

$\Rightarrow f(x) = (x^{2}+1)(x^{3}+9x^{2}+6x+7)=0$

Hence we have two roots of $f(x)=i,-i$

For the rest of the roots, we will solve :

$x^{3}+9x^{2}+6x+7 = 0$

In the next step of solving the cubic we will first depress the cubic( that is make an appropiate substitution such that co-efficient of $x^{2}=0$ )

In depressing the cubic $ax^{3}+bx^{2}+cx+d=0$ , we will make the substitution :

$x+\dfrac{b}{3a}=y$

In our cubic we make the substitution $x=y-3$ , to get our cubic as :

$y^{3}+43 = 21y$

Now we will make use of the identity :

$a^{3}+b^{3}+y^{3} - 3aby = (a+b+y)(a^{2}+b^{2}+y^{2}-ab-by-ay)$

Now we will find $a,b$ such that :

$a^{3}+b^{3}=43 , 3ab=21$

Eliminating $b$ from these two equations we have :

$a^{6}-43a^{3}+343=0$

It is quadratic in ${a}^{3}$ , hence solving for $a^{3}$ , we have ;

$a^{3} = \dfrac{43 \pm 3\sqrt{53}}{2}$

The real root of cubic in $y$ is given by :

$y = - a - b$

Since $(a+b+y)(a^{2}+b^{2}+y^{2}-ab-ay-by) = 0$

So let $a = \sqrt [ 3 ]{ \dfrac { 43-3\sqrt { 53 } }{ 2 } }$ , remember choice of $a$ is arbitary.

Hence $b=7\sqrt [ 3 ]{ \dfrac { 2 }{ 43-3\sqrt { 53 } } }$

Putting these values we have :

$y=- \sqrt [ 3 ]{ \dfrac { 43-3\sqrt { 53 } }{ 2 } } - 7\sqrt [ 3 ]{ \dfrac { 2 }{ 43-3\sqrt { 53 } } }$

Finally we have :

$x = -3 - \sqrt [ 3 ]{ \dfrac { 43-3\sqrt { 53 } }{ 2 } } - 7\sqrt [ 3 ]{ \dfrac { 2 }{ 43-3\sqrt { 53 } } }$

$\Rightarrow \boxed{a=3,b=43,c=53}$

$\text{Method 2}$

Put the equation into wolfram-alpha and press OK , and yes you got it.

I believe question poser has solved this question by this alternate way since as soon as I press OK , it returns me exactly the same closed form as he has posed in this problem.