How to go from $u^{2/3}-5u^{1/3}+7=0$ to $u^2-20u+343=0$

Question

I have absolutely no clue how to do this - any hints would be appreciated.

So far, I have tried multiplying each term by a power of u, but this leads nowhere...

Answer 1

Here is a solution inspired by the mechanical process to calculate a Groebner basis of the ideal $\langle x^2 - 5x + 7, x^3 - u \rangle$ in $\mathbb{Q}[x, u]$.

Let $x := u^{1/3}$; then we have the system of equations $$x^2 - 5x + 7 = 0, \quad (1) \\ x^3 - u = 0. \quad (2)$$ Now, let us try to eliminate $x$. By taking $x (1) - (2)$, we get: $$-5x^2 + 7x + u = 0. \quad (3)$$ Similarly, $5(1) + (3)$ now gives: $$-18x + 35 + u = 0. \quad (4)$$ Now, from here we can get $x = \frac{1}{18} (u + 35)$; then substituting into (1) will give a quadratic polynomial equation in terms of $u$. Using a CAS, I get that the substitution gives $\frac{1}{324} (u^2 - 20u + 343) = 0$.

Answer 2

Hint: write $t=u^{1/3}$ so $t^2-5t+7=0$. Plug now $u=t^3$ in

$$f(u) = u^3-20u+343 = t^9-20t^3+343 = ... $$ You must get $f(u)=0$. Don't forget to use $t^2 = 5t-7$

Answer 3

http://www.wolframalpha.com/input/?i=u%5E(2%2F3)+-+5u%5E(1%2F3)+%2B+7+%3D+0 http://www.wolframalpha.com/input/?i=u%C2%B3+-+20u+%2B+343+%3D+0

The solutions of both equations are not the same. Hence there is no rewriting the one into the other.

Answer 4

Hint:

$$(-7)^3=(u^{2/3}-5u^{1/3})^3=u^2-125u-15u(-7)$$

Answer 5

Cube both sides of $u^{2/3}=5u^{1/3}-7$ to get $$ \begin{align} u^2 &=125u\overbrace{-525u^{2/3}+735u^{1/3}}^{-105u}-343\\ &=20u-343 \end{align} $$ where $-105u=-105\overbrace{\left(5u^{1/3}-7\right)}^{u^{2/3}}u^{1/3}$

Therefore, $$ u^2-20u+343=0 $$

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In General

Suppose we had $u^{2/3}=au^{1/3}+b$, then $$ \begin{align} u &=au^{2/3}+bu^{1/3}\\ &=a\left(au^{1/3}+b\right)+bu^{1/3}\\ &=\color{#C00}{\left(a^2+b\right)u^{1/3}+ab}\\ u^2 &=\left(a^2+b\right)^2u^{2/3}+2ab\left(a^2+b\right)u^{1/3}+(ab)^2\\ &=\left(a^2+b\right)^2\left(au^{1/3}+b\right)+2ab\left(a^2+b\right)u^{1/3}+(ab)^2\\ &=\color{#C00}{\left(a^2+b\right)\left(a^3+3ab\right)u^{1/3}+b\left(a^4+3a^2b+b^2\right)} \end{align} $$ Then, we can combine the two red formulas to get $$ u^2=\left(a^3+3ab\right)u+b^3 $$

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Complex Approach

Let $v=u^{1/3}$ and $\omega=e^{2\pi i/3}$. Then note that $$ \left(v^2-av-b\right)\left(v^2\omega^2-av\omega-b\right)\left(v^2\omega-av\omega^2-b\right)=v^6-\left(a^3+3ab\right)v^3-b^3 $$