I am stuck on the following problem. I also mean to say greet you with a Hello! but for some reason it won't save it haha.
This problem was part of my homework, which is now past due and I have received partial credit for. The answer was released and I am trying to understand it.
I am completely stuck on the "Solving the equation $1+ \frac{2\sqrt{21}}{9} = (3a+a\sqrt{21})^3 $ for $a$" part. No matter how I break it down, I can't get that answer. Also, it seems like the $a=-6$ as the coefficient of the $x^2$ term has nothing to do with this part of the problem. Can someone explain please?
It is confusing for them to use $a$ in the hint, since it coincides with the coefficient of the $x^2$ term, and has nothing to do with the purpose of the hint.
Now the hint was
\begin{align}1+ \frac{2\sqrt{21}}{9} &= (3a+a\sqrt{21})^3 \\&=a^3(3+\sqrt{21})^3 \\&= a^3(3^3+3\cdot3^2\sqrt{21}+3\cdot3\sqrt{21}^2+\sqrt{21}^3) \\&=a^3(27+27\sqrt{21}+189+21\sqrt{21}) \\&=a^3(216+48\sqrt{21}) \\\leadsto a^3=\frac 19\frac{9+2\sqrt{21}}{216+48\sqrt{21}}&=\frac19\frac1{24}=\frac1{216}\end{align} (Factor the denominator and the numerator appears so a cancellation takes place)
which gives $a=\dfrac16$. Hence $1+ \dfrac{2\sqrt{21}}{9} = (3a+a\sqrt{21})^3 = \left(\dfrac12+\dfrac16\sqrt{21}\right)^3$.
This hint is only used to simplify the root
$$u=\sqrt[3]{1+ \frac{2\sqrt{21}}{9}}$$