Reciprocal equation problems: Difficult Problems with Solutions

Solution:
Dividing by [tex]x^2\ne 0[/tex] and substituting [tex]t=x+\dfrac{1}{x}[/tex] [tex]\left(x^2+\dfrac{1}{x^2}=t^2-2\right)[/tex]:
[tex]t^2-2t-8=0 \Rightarrow t=4, t=-2[/tex]
[tex]t=4:\ x^2-4x+1=0 \Rightarrow x=2\pm\sqrt{3}[/tex]
[tex]t=-2:\ x^2+2x+1=0 \Rightarrow x=-1[/tex]

Solution:
Dividing by [tex]x^2\ne 0[/tex] and substituting [tex]t=x+\dfrac{1}{x}[/tex] [tex]\left(x^2+\dfrac{1}{x^2}=t^2-2\right)[/tex]:
[tex]4\left(t^2-2\right)+12t-47=0 \iff 4t^2+12t-55=0 \Rightarrow t=\dfrac{5}{2}, t=-\dfrac{11}{2}[/tex]
[tex]t=\dfrac{5}{2}:\ 2x^2-5x+2=0 \Rightarrow x=2, x=\dfrac{1}{2}[/tex]
[tex]t=-\dfrac{11}{2}:\ 2x^2+11x+2=0 \Rightarrow x=\dfrac{-11\pm\sqrt{105}}{4}[/tex]

Solution:
Dividing by [tex]x^2\ne 0[/tex]: [tex]6\left(x^2+\dfrac{1}{x^2}\right)+7\left(x-\dfrac{1}{x}\right)-36=0[/tex]
Substitute [tex]t=x-\dfrac{1}{x}, x^2+\dfrac{1}{x^2}=t^2+2[/tex]: [tex]6t^2+7t-24=0 \Rightarrow t=\dfrac{3}{2}, t=-\dfrac{8}{3}[/tex]
[tex]t=\dfrac{3}{2}:\ 2x^2-3x-2=0 \Rightarrow x=2, x=-\dfrac{1}{2}[/tex]
[tex]t=-\dfrac{8}{3}:\ 3x^2+8x-3=0 \Rightarrow x=\dfrac{1}{3}, x=-3[/tex]

Solution:
[tex]P(x)=3\left(x^3+1\right)-ax(x+1)=(x+1)\left(3x^2-(3+a)x+3\right)[/tex]
So [tex]x=-1[/tex] is always a root. The quadratic factor is reciprocal (the product of its roots is [tex]1[/tex]), so its other root is [tex]-\dfrac{7}{2}[/tex].
[tex]x_1=-\dfrac{7}{2}, x_2=-1, x_3=-\dfrac{2}{7}[/tex]

Solution:
Dividing by [tex]t^2\ne 0[/tex]; Substitute [tex]y=t+\dfrac{1}{t}, t^2+\dfrac{1}{t^2}=y^2-2[/tex]: [tex]y^2-2+2y+1=0 \iff y^2+2y-1=0 \Rightarrow y=-1\pm\sqrt{2}[/tex]
[tex]y=-1+\sqrt{2}:\ |y|\lt 2[/tex] — no real roots.
[tex]y=-1-\sqrt{2}:\ t^2+\left(1+\sqrt{2}\right)t+1=0, D=\left(1+\sqrt{2}\right)^2-4=2\sqrt{2}-1\gt 0[/tex]
[tex]t_{1,2}=\dfrac{-1-\sqrt{2}\pm\sqrt{2\sqrt{2}-1}}{2}[/tex]