Mathematical reasoning

Question

Given $x = ( 19-8^{1/3})^{1/2}$

We need to find the value of $$\frac{x^4 - 6x^3 - 2x^2 + 18x + 15}{x^2 - 8x + 15}$$

I found out $x^2 - 8x + 13=0$ and then attempted to put this to use in the expression for which I got the answer as $1$. I am not sure about it. You all can also try. It's fun doing it.

Answer 1

We have $$ x = (19-8^{1/3})^{1/2}=\sqrt{17}$$ so $$\begin{align}\frac{x^4 - 6x^3 - 2x^2 + 18x + 15}{x^2 - 8x + 15}&=\frac{289-102\sqrt{17}-34+18\sqrt{17}+15}{17-8\sqrt{17}+15}\\&=\frac{270-84\sqrt{17}}{32-8\sqrt{17}}\\&=\frac{270-84\sqrt{17}}{32-8\sqrt{17}}\cdot\frac{32+8\sqrt{17}}{32+8\sqrt{17}}\\&=\frac{-2784-528\sqrt{17}}{-64}\\&=\frac{174+33\sqrt{17}}4\end{align}$$

Answer 2

HINT:

$$\frac{x^4-6x^3-2x^2+18x+15}{x^2-8x+15}=x(2+x)+\frac{15}{x-3}-\frac{35}{x-5}-1\tag1$$

So, in your case:

$$x(2+x)=17+2\sqrt{17}\tag2$$