Show that $a^3+5a$ is an integer

Question

I've been given the following task. Let

$$a = \sqrt[3]{1+\sqrt{\frac{152}{27}}}-\sqrt[3]{-1+\sqrt{\frac{152}{27}}}$$

Show that $a^3+5a$ is an integer.

I tried calculating it by hand but the small page of my copybook is not large enough for the very long calculations. Is there a trick I could use here instead of calculating it by hand?

I tried factoring $a^3+5a$ into $a(a^2+5)$ to make it more simple to calculate but it still gets a tad bit complicated when multiplying with $a$ again.

Thank you in advance.

Answer 1

Let $x=\sqrt[3]{1+\sqrt{\frac{152}{27}}}$ and $y=\sqrt[3]{-1+\sqrt{\frac{152}{27}}}$ .

Then $$\begin{aligned}a^3+5a&=(x-y)^3+5(x-y)\\ &=x^3-3xy(x-y)-y^3+5(x-y)\\ &=x^3-y^3+(5-3xy)(x-y) \end{aligned}$$

Now,

$\begin{aligned}5-3xy&=5-3\sqrt[3]{\left(1+\sqrt{\frac{152}{27}}\right)\left(-1+\sqrt{\frac{152}{27}}\right)}\\ &=5-3\sqrt[3]{\frac{125}{27}}\\ &=5-3\cdot\frac53\\ &=0 \end{aligned}$

and $x^3-y^3=2$, so $a^3+5a=2$.

Answer 2

Let $x=1+\sqrt{\frac{152}{27}},~~y=-1+\sqrt{\frac{152}{27}}$

$$x^{\frac{1}3}y^{\frac{1}3}=\frac{5}{3}$$

$$a=x^{\frac{1}3}-y^{\frac{1}3}$$ $$\begin{align} a\left(x^{\frac{2}3}+x^{\frac{1}3}y^{\frac{1}3}+y^{\frac{2}3}\right)&=\left(x^{\frac{1}3}-y^{\frac{1}3}\right)\left(x^{\frac{2}3}+x^{\frac{1}3}y^{\frac{1}3}+y^{\frac{2}3}\right)\\ \\ a\left(x^{\frac{2}3}+x^{\frac{1}3}y^{\frac{1}3}+y^{\frac{2}3}\right)&=x-y=2\\ \\ a\left(\left(x^{\frac{1}3}-y^{\frac{1}3}\right)^2+3x^{\frac{1}3}y^{\frac{1}3}\right)&=2\\ \\ a\left(a^2+5\right)&=2\\ \\ a^3+5a&=2 \end{align}$$

Answer 3

From the Binomial Theirem:

$(u+v)^3=u^3+3u^2v+3uv^2+v^3=(u^3+v^3)+3uv(u+v).$

So with

$u=\sqrt[3]{1+\sqrt{\dfrac{152}{27}}}$

$v=\sqrt[3]{1-\sqrt{\dfrac{152}{27}}}$

(where in the expression for $v$ we have used $-\sqrt[3]{u}=\sqrt[3]{-u}$), we have $a=u+v$ and then

$a^3=\color{blue}{\left(1+\sqrt{\dfrac{152}{27}}\right)+\left(1-\sqrt{\dfrac{152}{27}}\right)}+\color{brown}{3\left(\sqrt[3]{1+\sqrt{\dfrac{152}{27}}}\right)\left(\sqrt[3]{1-\sqrt{\dfrac{152}{27}}}\right)}(a).$

The blue terms simplify to $2$. With the difference of squares factorization the brown terms simplify to $-3\sqrt[3]{125/27}=-5$. Therefore

$a^3=2-5a.$

This is essentially the reverse of the Cardano method for solving cubic equations.