Obvious way to solve it is to substitute $y^2=(x-1)^3$ right away. However, after that I am stuck with solving equation $2x^3 + 3(x-1)^5=0$, which for some reason occurred difficult to me.

So, I thought that maybe using Lagrange method could help. But in fact, in this case it led me to the same equation.

So, my questions are:

  1. Is there any simple way to solve this equation?
  2. Is it a coincidence that in this case Lagrange method didn’t give me anything useful? Or I could see this in advance?

Thanks for your help and answers in advance!

  • 3
    $\begingroup$ Actually, I think I figured it out myself. The thing is that $(x-1)^3=y^2 \ge 0$ which follows that $(x-1)^3 \ge 0$, hence we know that $x$ should be at least 1. And for polynomial that I mentioned it is correct that $P(x) \ge 0$ for $x \ge 1$. So, function grows on $[1, \infty)$, hence minimum is in $x=1$, u = 1. $\endgroup$ Oct 13 at 23:28
  • 1
    $\begingroup$ Commenting, rather than answering, because your comment nailed it. Reactions: [1] In general, analysis such as in your comment, belongs in the posted question itself, not a comment. You can simply add an Edit or Addendum section to your answer. [2] My approach would have been that $y^2 = (x-1)^3 \implies y^4 = (x - 1)^6.$ So, $u = x^4 + (x-1)^6.$ Then, I would have let $t = (x-1) \implies u = (t+1)^4 + t^6 \implies u' = 4(t+1)^3 + 6t^5.$ As you indicated, $t < 0$ is disallowed, and for $t > 0$, you have that $u'$ is always positive. $\endgroup$ Oct 13 at 23:47

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