I need to solve the following with symmetric polynomials, but I do not understand what to do here. $$\sqrt[3]{\sqrt {x-1} - 2} = 1-\sqrt[3]{9-\sqrt{x-1}}$$
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2Your question might be closed unless you show some work toward solving it yourself. You should edit it to show some of your work. HINT: Can you find two equations in $u$ and $v$ if you let $u^3=\sqrt{x-1}-2$ and $v^3=9-\sqrt{x-1}$? Then see if you can replace the original equation with an equation with just one variable, either $u$ or $v$. – John Wayland Bales Dec 22 '21 at 19:13
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I can`t understand. I got u = 1-v, but I do not understand what to do next. – padalad499 Dec 22 '21 at 19:29
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What does $u^3+v^3$ equal. That would give you a second equation in $u$ and $v$ making it possible to solve for $u$ and $v$ and then for $x$. – John Wayland Bales Dec 22 '21 at 19:32
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And you should add what you get to your question so your work can be seen. Questions which show no partial work are frequently closed or downvoted. – John Wayland Bales Dec 22 '21 at 19:34
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Thank you for hints! I want to try to solve it myself now and then I`ll show my solution. – padalad499 Dec 22 '21 at 19:35
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Excellent idea! – John Wayland Bales Dec 22 '21 at 19:36
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Thank you very much! I solve it thanks to your hints! – padalad499 Dec 22 '21 at 19:48
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Yes, that is the correct solution. Congratulations! But you should keep the original post and just add the graphic to it. – John Wayland Bales Dec 22 '21 at 19:49
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Recheck the solution to your equation $\sqrt{x-1}=1$. – John Wayland Bales Dec 22 '21 at 19:51
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There has to be a square root. – padalad499 Dec 22 '21 at 19:54
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Yes, I corrrected it. – John Wayland Bales Dec 22 '21 at 19:54
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Everything is OK, I think. – padalad499 Dec 22 '21 at 19:58