Volume of cube with part of edge removed.

Hello, domenfrandolic!

Quote:

$\text{The diagram below shows a solid with volume }V,$
$\text{obtained from a cube with edge }a > 1$
$\text{ with a smaller cube with edge }\tfrac{1}{a}\text{ removed.}$

$\text{Let }x\:=\:a - \tfrac{1}{a}$

$\text{(a) Find }V\text{ in terms of }x.$

$\text{(b) Hence or otherwise, show that the only value of }a$
. . . . $\text{ for which }V\,=\,4x\,\text{ is: }\: a\:=\: \tfrac{1+\sqrt{5}}{2}$

(a) We have: . $x \:=\:a - \frac{1}{a}$ .[1]

Square: . $x^2 \:=\:\left(a-\frac{1}{a}\right)^2 \:=\:a^2-2+\frac{1}{a^2} \quad\Rightarrow\quad a^2 + \frac{1}{a^2} \:=\:x^2+2$ .[2]

The volume is: . $V \;=\;a^3 - \left(\frac{1}{a}\right)^3 \;=\;a^3 - \frac{1}{a^3}$

Factor: . $V \;=\;\left(a - \frac{1}{a}\right)\left(a^2 + 1 + \frac{1}{a^2}\right) \;=\;\underbrace{\left(a - \frac{1}{a}\right)} \left(\underbrace{a^2+\frac{1}{a^2}} + 1\right)$

Substitute [1] and [2]: . $V \;=\;(x)\,(x^2+2 + 1) \quad\Rightarrow\quad \boxed{V \;=\;x(x^2+3)}$

$\text{(b) If }V \,=\,4x,\text{ we have:}$

. . $a^3 - \frac{1}{a^3} \;=\;4\left(a-\frac{1}{a}\right) \quad\Rightarrow\quad \left(a^3-\frac{1}{a^3}\right) - 4\left(a-\frac{1}{a}\right) \;=\;0$

Factor: . $\left(a-\frac{1}{a}\right)\left(a^2 + 1 + \frac{1}{a^2}\right) - 4\left(a - \frac{1}{a}\right) \;=\;0$

Factor: . $\left(a - \frac{1}{a}\right)\left(a^2 + 1 + \frac{1}{a^2} - 4\right) \;=\;0$

. . . . . . . . . $\left(a - \frac{1}{a}\right)\left(a^2 - 3 +\frac{1}{a^2}\right) \;=\;0$

We have two equations to solve . . .

$a - \frac{1}{a} \:=\:0 \quad\Rightarrow\quad a^2 - 1 \:=\:0 \quad\Rightarrow\quad \rlap{///////}a \:=\:\pm1 \;\;\text{ (Recall that }a > 1)$

$a^2 - 3 + \frac{1}{a^2} \;=\;0 \quad\Rightarrow\quad a^4 - 3a^2 + 1 \;=\;0$
. . $\text{Quadratic Formula: }\;a^2 \:=\:\frac{3\pm\sqrt{5}}{2}\;=\;\left(\frac{1\pm \sqrt{5}}{2}\right)^2$
. . $\text{Therefore: }\;a\;=\;\frac{1 + \sqrt{5}}{2}$