## Task B-2.3.

From a cube with edge length $a>1$, a cube with edge length $\frac{1}{a}$ has been cut out at one of its vertices. Determine $a$ such that the volume of the newly formed body is $V=4\left(a-\frac{1}{a}\right)$.
## First Solution.

According to the conditions of the problem, we set up the equation $\quad a^{3}-\frac{1}{a^{3}}=4\left(a-\frac{1}{a}\right)$.

After simplifying, we get $\quad a^{6}-4 a^{4}+4 a^{2}-1=0$.

If we factorize the left side, we group the first and last term and the second and third term. Then we have

$$
\begin{gathered}
a^{6}-1-4 a^{2}\left(a^{2}-1\right)=0 \\
\left(a^{2}-1\right)\left(a^{4}+a^{2}+1\right)-4 a^{2}\left(a^{2}-1\right)=0 \\
\left(a^{2}-1\right)\left(a^{4}-3 a^{2}+1\right)=0
\end{gathered}
$$

Since $a>1$, it follows that $a^{2}-1 \neq 0$, so we have

$a^{4}-3 a^{2}+1=0$,

from which $a^{2}=\frac{3+\sqrt{5}}{2}$, or $a=\sqrt{\frac{3+\sqrt{5}}{2}}$.

1 point

Note: It holds that $\frac{3+\sqrt{5}}{2}=\left(\frac{1+\sqrt{5}}{2}\right)^{2}$, so $a=\frac{1+\sqrt{5}}{2}$, but the student can leave the expression as $a=\sqrt{\frac{3+\sqrt{5}}{2}}$.