Simplify the following expression!

Question

Simplify the following expression

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

I tried using the form $\displaystyle\frac{a^3+b^3}{a^2-ab+b^2}$, and i also tried to assume the requested expression with $c$ and use the form $a^3+b^3-c^3=-3abc$, if $a+b-c= 0$, but haven't obtained its simple form yet,

i get this problem from a book but i doesn't get the solution

if anyone can complete it it's really amazing

Answer 1

A first (direct) approach:

Formula :

$$E:=\sqrt[3]{a+\frac{a+8}{3}\sqrt{\frac{a-1}{3}}}+\sqrt[3]{a-\frac{a+8}{3}\sqrt{\frac{a-1}{3}}},\tag{1}$$

contains expression

$$u:=\sqrt{\frac{a-1}{3}}\tag{2}$$

which is only meaningful under condition $a \ge 1$.

We set apart the case $a=1$ where $E=2$.

Let us invert (2) under the form

$$a=3u^2+1\tag{3}$$

Then:

$$\frac{a+8}{3}=\frac{3u^2+9}{3}=u^2+3$$

In this way, (1) becomes:

$$E=\sqrt[3]{(3u^2+1)+(u^2+3)u}+\sqrt[3]{(3u^2+1)-(u^2+3)u}$$

$$E=\sqrt[3]{(u+1)^3}+\sqrt[3]{-(u-1)^3}$$

$$E=u+1-(u-1) = \color{red}{2}$$

which is a constant independent from $a$.

A second approach:

Formula (1) has the same "structure" as Cardano formula, expressing a root of the general reduced cubic equation:

$$x^3+px+q=0\tag{4}$$

under the form

$$\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{D}{4 \times 27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{D}{4 \times 27}}} \ \text{where} \ D:=4p^3+27q^2,\tag{5}$$

($D$ is the so-called discriminant of equation (4)).

Identifying (1) and (5), one gets:

$$a=-\frac{q}{2} \tag{6}$$

and:

$$\frac{(a+8)^2(a-1)}{27}=\frac{4p^3+27q^2}{4 \times 27} \tag{7}$$

Plugging (6) into (7), we get:

$$(a+8)^2(a-1)=p^3+27a^2\tag{8}$$

Expanding (8), one gets:

$$p^3=a^3-12a^2+48a-64=(a-4)^3 \ \ \implies \ \ p=a-4\tag{9}$$

Plugging (6) and (9) into (4):

$$x^3+(a-4)x-2a=0 \ \ \iff \ \ (x-2)(x^2+2x+a)=0$$

with roots:

$$\begin{cases}x_1&=&\color{red}{2}\\ x_2&=&-1-\sqrt{1-a^2}\\ x_3&=&-1+\sqrt{1-a^2}\end{cases}$$

We find back the solution $\color{red}{x=2}$.

The two other solutions $x_2, x_3$ are in fact complex in general because $1-a^2<0$ which is not possible because expression $E$ in (1) is assumed to be real.

Remark: Here is a formula with 2 variables with a complicated LHS, similar to (1) and a very simple RHS, free from one of the two variables:

$$\sqrt[3]{n\left(3m-n^2\right) +m\sqrt{8m-3n^2}} + \sqrt[3]{n\left(3m-n^2\right) -m\sqrt{8m-3n^2}} =n$$

Answer 2

This answer was started some longer time ago, now i finally had time to complete it. Although we already have a good, compact, accepted answer i will still post it since the path of getting the solution is (slightly) different, and may be of interest.

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Let $u,v$ be the two numbers $$ \sqrt[3]{a\pm\frac {a+8}3\sqrt{\frac{a-1}3}}\ . $$ Then we have $u^3+v^3=2a$ and $$ \begin{aligned} (uv)^3 &=u^3v^3 \\ &= \left(a+\frac {a+8}3\sqrt{\frac{a-1}3}\right) \left(a-\frac {a+8}3\sqrt{\frac{a-1}3}\right) \\ &=\frac 1{27}\left(27a^2-(a+8)^2(a-1)\right) \\ &=-\frac 1{27}(a-4)^3 \ . \end{aligned} $$ Assuming that $u,v\in\Bbb R$, and this happens in the case and only in the case of $\sqrt{(a-1)/3}\in\Bbb R$, i.e. $a\ge 1$, (and only when we agree to take two real values in $\sqrt[3]{\dots}$) we obtain $$uv=-\frac 13(a-4)\ .$$ This implies: $$ \begin{aligned} (u+v)^3 &= (u^3+v^3)+3uv(u+v)\\ &=2a - (a-4)(u+v)\ , \end{aligned} $$ so $u+v$ is a root of the polynomial of degree three $X^3 + (a-4)X -2a=(X-2)(X^2+2X+a)$.

• For $a=1$ we easily check $u=v=1$, so $u+v=2$. Else...
• For $a>1$ we can (and as assumed do) perform all computations with $u,v\in\Bbb R$, so $(u+v)$ is the only real solution of $(X-2)(X^2+2X+a)$, which is $2$.
• For $a<1$ or for a general complex $a$ we still can build $u,v\in\Bbb C$. Assuming that we take the two roots $\sqrt[3]{\dots}$ in $u,v$ in $\Bbb C$ so that we still have the compatibility $uv=-(a-4)/3$ and we obtain at any rate a root of the mentioned polynomial for the value $u+v$.