College Algebra, Chapter 4, 4.4, Section 4.4, Problem 102

Suppose that a rectangular box with a volume of $2\sqrt{2}\text{ft}^3$ has a square base as shown in the figure. The diagonal
of the box is $1\text{ft}$ longer than each side of the base.
a.) If the base has sides of length $x$ feet, show that $x^6 - 2x^5 - x^4 + 8 = 0$
b.) Show that two different boxes satisfy the given conditions. Find the dimensions in each case.

a.) If the value of the box is $2\sqrt{2}\text{ft}^3$, then
$2\sqrt{2}=x^2 y$ Equation 1
Hence, by using Pythagorean Theorem,



$
\begin{equation}
\begin{aligned}
y^2 + (x\sqrt{2})^2 &= (x+1)^2\\
\\
y^2 + 2x^2 &= (x+1)^2\\
\\
y^2 &= (x+1)^2 - 2x^2\\
\\
y &= \sqrt{(x+1)^2 - 2x^2} && \text{Equation 2}
\end{aligned}
\end{equation}
$


By substituting Equation 2 and Equation 1, we get

$
\begin{equation}
\begin{aligned}
2\sqrt{2} &= x^2 \left( \sqrt{(x+1)^2 - 2x^2 }\right) && \text{Model}\\
\\
8 &= x^4 \left[ (x+1)^2 - 2x^2 \right] && \text{Square both sides}\\
\\
8 &= x^4 \left[ x^2 + 2x+1 -2x^2 \right] && \text{Expand}\\
\\
8 &= x^4 \left[ -x^2 + 2x + 1 \right] && \text{Combine like terms}\\
\\
8 &= -x^6 + 2x^5 + x^4
\end{aligned}
\end{equation}
$

Thus,
$x^6 - 2x^5 - x^4 + 8 = 0$