$$E_{1} : \sqrt[3]{1+z}-\sqrt[3]{1-z}=\sqrt[6]{1-z^{2}} $$
Let $a=\sqrt[3]{1+z}$ and $b=\sqrt[3]{1-z}$
$E_1$ is equivalent to $E_2$ :
$$ E_2:\ \dfrac{\sqrt{a}}{\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}}=1$$
Let $t=\dfrac{\sqrt{a}}{\sqrt{b}}$, and $t$ verify :
$$ E_{3}:\ t^2-t-1 $$
- My Question :
Solve the equation $E_{3}$ and deduce the solution $\alpha$ to the equation $E_{1}$ ?
Indeed,
the discriminant of $E_3$ is $\Delta t=5\geq 0$ then there are two distinct roots
$t=\dfrac{1\pm \sqrt{5}}{2} $ or $t=\dfrac{\sqrt{a}}{\sqrt{b}}$
$\sqrt{a}=t\sqrt{b} \Longleftrightarrow a=t^2b$
$(t_1, t_2) \in \{\dfrac{1\pm \sqrt{5}}{2}\} \Longleftrightarrow S:\ \begin{cases} a=t_1^2b \\ \\ a=t_2^2b \end{cases} $
and by WF give us for $E_{1}$ one solution
$z = 2/\sqrt{5}$
I'm stuck here i can't solve S
- Is my reasoning correct ? Is There simple way to solve the $E_{1}$
Thanks for any help.
Your calculations are okay. To finish it off you can use the relation $a^3 + b^3 = 2$ which comes from $E1$. From this we have $(a/b)^3 = (2/b)^3 - 1$ (note $b\neq 0$). The left hand side is $t^6$, so $1+t^6 = (2/b)^3$, or $b^3 = \frac{8}{1+t^6}$. Since $z=b^3-1$ we have $z = \frac{8}{1+t^6}-1$.
Now some long division of polynomials will give $$t^6 = (t^4+t^3+2t^2+3t+5)(t^2-t-1)+8t+5 = 8t+5,$$ so in fact $z=\frac{8}{8t+6}-1 = \frac{8}{10\pm4\sqrt{5}}-1 = \frac{8(10 \mp 4 \sqrt{5})}{100-20}-1 = \frac{10\mp 4 \sqrt{5}}{10} - 1 = \mp \frac{2}{\sqrt{5}}$.
Since $-\frac{2}{\sqrt{5}}$ doesn't satisfy $E_1$ (makes the left hand side negative), we have $z=\frac{2}{\sqrt{5}}$.