[tex]16(x^3+1)^2-22(x^3+1)-3=0[/tex]
Use substitution: [tex]x^3+1=t[/tex]
[tex]16t^2-22t-3=0\\\\16t^2-24t+2t-3=0\\\\8t(2t-3)+1(2t-3)=0\\\\(2t-3)(8t+1)=0\iff2t-3=0\ \vee\ 8t+1=0[/tex]
[tex]2t-3=0\ \ \ |+3\\\\2t=3\ \ \ |:2\\\\t=\dfrac{3}{2}\\..............................\\8t+1=0\ \ \ \ |-1\\\\8t=-1\ \ \ \ |:8\\\\t=-\dfrac{1}{8}[/tex]
we're going back to substitution:
[tex]x^3+1=\dfrac{3}{2}\ \vee\ x^3+1=-\dfrac{1}{8}\ \ \ \ |subtract\ 1\ from\ both\ sides\ of\ the\ equations\\\\x^3=\dfrac{1}{2}\ \vee\ x^3=-\dfrac{9}{8}\\\\x=\sqrt[3]{\dfrac{1}{2}}\ \vee\ x=\sqrt[3]{-\dfrac{9}{8}}[/tex]
[tex]x=\dfrac{1}{\sqrt[3]2}\ \vee\ x=-\dfrac{\sqrt[3]9}{2}\\\\\boxed{x=\dfrac{\sqrt[3]4}{2}\ \vee\ x=-\dfrac{\sqrt[3]9}{2}}[/tex]
