How to simplify $\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}$?

Question

How to simplify this? $$\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}$$

Answer 1

after the hint above we get $$\left(\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\left(\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}\right)\right)\left(\left(\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}\right)^2-\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\left(\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}\right)^2\right)=a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}+a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}$$ can you simplify this?

Answer 2

$s=\sqrt[3]{a+\frac{a+1}3\sqrt{\frac{8a-1}3}}=\sqrt[3]{a+b}$

$t=\sqrt[3]{a-\frac{a+1}3\sqrt{\frac{8a-1}3}}=\sqrt[3]{a-b}$

$\bbox[5px,border:1px solid]{s^3+t^3=2a}$

$\displaystyle (st)^3=(a+b)(a-b)=a^2-b^2=a^2-\frac{(a+1)^2(8a-1)}{27}=-\frac {(2a-1)^3}{27}$

$\bbox[5px,border:1px solid]{-3st=(2a-1)j^k}\quad$ with $j^3=1$ and $k=0,1,2$

$(s+t)^3=(s^3+3s^2t+3st^2+t^3)=(s^3+t^3)+3st(s+t)$

So if we set $X=s+t$ then $X$ is solution of $X^3-3stX-(s^3+t^3)=0$

$\bbox[5px,border:1px solid]{X^3+(2a-1)j^kX-2a=0}$

The case $k=0$ is simple :

$X^3-X+2a(X-1)=X(X-1)(X+1)+2a(X-1)=(X-1)(X^2+X+2a)=0$

$X=1$ or $\displaystyle X=\frac{-1\pm\sqrt{1-8a}}2$

According to the graph in real values:

It seems $X=1$ for $a\in[-1,\frac 12]$ and $X=\frac{-1-\sqrt{1-8x}}2$ for $a<-1$.

However I suspect the grapher to take particular branches of $\sqrt[3]{}$.

In fact among the $9$ possible values for $s+t$ (three roots for $s$ and same for $t$ gives $9$ possibilities for $s+t$), I think that $1,j,j^2$ are always reachable by some combination. When trying manually with the help of a CAS, $s,t=(\frac 12\pm x\sqrt{y})j^k$ so this is possible to get $1=\frac 12+\frac 12$ and the other values I get for instance with the green curve above, is just $s+t$ with a $k_s$ and $k_t\neq 0$.

But I have no idea how to present a clean argument for this observation.

Answer 3

$s=\sqrt[3]{a+\frac{a+1}3\sqrt{\frac{8a-1}3}}$

$t=\sqrt[3]{a-\frac{a+1}3\sqrt{\frac{8a-1}3}}$

Let's have $\begin{cases} \displaystyle X=\frac 12\pm\frac 12\sqrt{\frac{8a-1}3}=x\pm x\sqrt{y}\\\\ \displaystyle x=\frac 12,\quad\displaystyle y=\frac{8a-1}{3}\end{cases}$

$\displaystyle X^3=(x^3\pm 3x^3\sqrt{y}+3x^3y\pm x^3y\sqrt{y})=x^3(3y+1)\pm x^3(3+y)\sqrt{y}$

$\displaystyle X^3=\frac 18(8a)\pm \frac 18\bigg(\frac{9+8a-1}3\bigg)\sqrt{y}=a\pm\frac{a+1}3\sqrt{\frac{8a-1}3}$

Thus the complex roots for $s,t$ are $Xj^k$ with $j^3=1$ and $k=0,1,2$.

Note that if $8a-1<0$ then the complex part of $(s+t)$ which resides in $\sqrt{y}$ cancels if we choose the same $k$ for $s$ and $t$.

In such a case, we have $s+t=2xj^k=j^k$ and for $k=0$ the special value $\bbox[5px,border:2px solid]{s+t=1}$

If we want to be complete there are in fact $9$ possible values :

$\displaystyle X(k,\sigma)=j^k\bigg(\frac 12+\sigma\frac 12\sqrt{\frac{8a-1}3}\ \bigg)$

Then the $9$ values are $\{X(k_1,1)+X(k_2,-1)\mid 0\le k_1,k_2 \le 2\}$

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Edit: a note on how I found $X$.

I had a doubt on my previous answer, and starting playing with Maple, I figured out that roots of $s,t$ seems to share this form $(\frac 12\pm x\sqrt{y})j^k$.

So I calculated $(\frac 12+x\sqrt{y})^3$ and by identification I found the constraint $\bbox[5px,border:1px solid]{12xy^2=8a-1}$

From there, I went on redacting this actual answer.

Answer 4

Let $$x=\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}$$ Take the cubic power of both sides $$x^3=2a - (2a-1)x$$ or $$(x-1)(x^2+x+2a)=0$$ Thus $$x=1=\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}$$