Minimalism is a decision - (1) - A different problem

Given $\begin{cases} x^2-5y + 1 = 0 & ...(1) \\ x - \sqrt y - 1 = 0 & ...(2) \end{cases}$ .

From $(2):$

$\begin{aligned} x - \sqrt y - 1 & = 0 & \small \color{#3D99F6} \text{Rearranging} \\ \sqrt y & = x-1 & \small \color{#3D99F6} \text{Squaring both sides} \\ y & = x^2 - 2x + 1 \end{aligned}$

From $(1):$

$\begin{aligned} x - 5{\color{#3D99F6}y} + 1 & = 0 & \small \color{#3D99F6} \text{Substituting } y = x^2 - 2x + 1 \\ x - 5{\color{#3D99F6}(x^2-2x+1)} + 1 & = 0 \\ - 4x^2 +10x - 4 & = 0 & \small \color{#3D99F6} \text{Dividing both sides by }-2 \\ 2x^2 - 5x + 2 & = 0 \\ (2x-1)(x-2) & = 0 \end{aligned}$

Then we have $\begin{cases} x = \frac 12 & \implies \sqrt y = x - 1 = -\frac 12 & \small \color{#D61F06}\text{No real solution.} \\ x = 2 & \implies \sqrt y = x - 1 = 1 & \implies y = 1 \end{cases}$

Therefore, $\sqrt[3]{2x^4-5y} = \sqrt[3]{2(16)-5} = \sqrt[3]{27} = \boxed 3$ .

By multiplying the second expression by -5 and adding it to the first, we obtain:

$x^2 - 5x + 6 = 0 \Rightarrow (x-2)(x-3) = 0 \Rightarrow x = 2, 3$

and substituting these values into either equation yields the ordered-pairs: $(x,y) = (2,1); (3,4).$ A final calculation of $(2x^4 - 5y)^{1/3}$ gives:

$(2(2)^4 - 5(1))^{1/3} = (27)^{1/3} = 3$

$(2(3)^4 - 5(4))^{1/3} = (142)^{1/3} = 5.216$

Hence, the answer is $\boxed{3}.$