If a and b are the distinct roots of the equation $x^2+3^{1/4}x + 3^{1/2} =0$ then find the value of $a^{96}(a^{12}-1)+b^{96}(b^{12}-1)$. 
If $a$ and $b$ are the distinct roots of the equation $x^2+3^{1/4}x +
 3^{1/2} =0$ then find the value of $a^{96}(a^{12}-1)+b^{96}(b^{12}-1)$

My attempt:
LHS = $(a^{108}+b^{108})-(a^{96}+b^{96})$
$a^{n} + b^{n} = -[3^{1/4}(a^{n-1}+b^{n-1})+3^{1/2}(a^{n-2}+b^{n-2})]$
$a^{108} + b^{108} = -[(3^{1/4}(a^{107}+b^{107})+3^{1/2}(a^{106}+b^{106})]$
I suppose i could repeat this process until every term in is terms of $a^{96}$ and $b^{96}$ but that would be very tedious
 A: As suggested in Eric Snyder's comment, let $c = \sqrt[4]{3}$. Then rearrange your equation to get
$$x^2 = -cx - c^2 \tag{1}\label{eq1A}$$
Multiply both sides by $x$ and use \eqref{eq1A} to get
$$\begin{equation}\begin{aligned}
x^3 & = -cx^2 - c^{2}x \\
& = -c(-cx - c^2) - c^{2}x \\
& = c^{2}x + c^3 - c^{2}x \\
& = c^3
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Thus, with $x = a$ and $x = b$ being the $2$ distinct roots of the equation in \eqref{eq1A}, we get
$$a^3 = b^3 = c^3 = (\sqrt[4]{3})^3 \; \; \to \; \; a^{12} = b^{12} = 3^3 = 27 \tag{3}\label{eq3A}$$
This results in
$$a^{96}(a^{12}-1)+b^{96}(b^{12}-1) = 2(27^{8})(27 - 1) = 52(3^{24}) \tag{4}\label{eq4A}$$

Alternatively, note $(x-c)(x^2+cx+c^2)=x^3-c^3=0 \; \to \; a^3=b^3=c^3$. Also, we could've used the quadratic formula with \eqref{eq1A} to get $a,b = c\left(\frac{-1\pm\sqrt{3}i}{2}\right)$. Cubing this (instead of doing the explicit calculations, it's simpler to use that $\frac{-1\pm\sqrt{3}i}{2}$ are third roots of $1$) then also gives $a^3 = b^3 = c^3$, as previously shown in \eqref{eq3A}.
A: $x^2+3^{\frac{1}{4}}x + 3^{\frac{1}{2}} =0$
$\implies \left(x^2+3^{\frac{1}{2}}\right)^2=\left(-3^{\frac{1}{4}}x\right)^2$
$\implies x^4+2x^2{3}^{\frac{1}{2}}+3=3^{\frac{1}{2}}x^2$
$\implies\left(x^4+3\right)^2=3x^4$
$\implies x^8+6x^4+9=3x^4$
$\implies x^8=-3x^4-9$
$ \begin{align}\implies x^{12}&=-3x^8-9x^4\\&=9x^4+27-9x^4\\&=27\end{align}$
Hence $a^{12}=27$ and $b^{12}=27$

$\begin{align}&a^{96}\left(a^{12}-1\right)+b^{96}\left(b^{12}-1\right)\\&=26\cdot 27^{8}+26\cdot27^{8}\\&=52\cdot 3^{24}\end{align}$

