For question 1

Let \(y=x^2-4x\)

Let \(x=y^2-4y\)

With this \(x \neq y\)

We want to know what \(x^2+y^2\) so we square both of the equations to get that

\(x^2=x^4-8x^3+16x^2\) and with \(y \) we get

\(y^2=y^4-8y^3+16y^2\) Since there are differerent coefficents on both \(x\) and \(y\)

All we can do is put it in biggest degree to lowest degree

\(x^2+y^2=x^4+y^4−8x^3−8y^3+16x^2+16y^2\)

Question 2. \(4=a+\frac{1}{a}\) multiply both sides by a

\(4a=a^2+1\) as you notice this a quadratic in disguise! hence

\(-a^2+4a-1=0\) as factorisation dosn't work we will use the quadratic formula

For values a=-1 b=4 c=-1

\(a = {-4 \pm \sqrt{4^2-4} \over -2}\)simplify \(a = {2 \pm \sqrt{12} }\)

Simplify even more \(a=2 \pm 2\sqrt{3}\)

Now to find the value of \(a^4+a^{-4}\)\(a^4=97\pm56\sqrt{3}\)\(a^{-4}=\frac{7}{64}\pm \frac{1}{16}\sqrt{3}\)

\(a^4 + a^{-4}= 97 \pm 56\sqrt{3} + \frac{7}{64} \pm \frac{1}{16}\sqrt{3}\) after a bit more simplifying we get

\(\frac{6215}{64}-\frac{897}{16}\sqrt{3}\) if \(97 - 56\sqrt{3} + \frac{7}{64} - \frac{1}{16}\sqrt{3}\) if \(97 + 56\sqrt{3} + \frac{7}{64} + \frac{1}{16}\sqrt{3}\) we get \(\frac{6215}{64}+\frac{897}{16}\sqrt{3}\) if \(97 + 56\sqrt{3} + \frac{7}{64} - \frac{1}{16}\sqrt{3}\) we get\(\frac{6215}{64}+\frac{895}{16}\sqrt{3}\) if

\(97 - 56\sqrt{3} + \frac{7}{64} + \frac{1}{16}\sqrt{3}\) we get \(\frac{6215}{64}-\frac{895}{16}\sqrt{3}\) that's all the combinations! PHEW...