The numbers $a, \; b, \; c \;$ and $\; d$ $\;$ form a geometric progression (G.P).
Find $\;$ $\left(a - c\right)^2 + \left(b - c\right)^2 + \left(b - d\right)^2 - \left(a - d\right)^2$
Since the numbers $a$, $b$, $c$ and $d$ form a G.P
$\implies$ $\dfrac{b}{a} = \dfrac{c}{b} = \dfrac{d}{c} = k$ $\;\;$ where $k$ is a constant
Now, $\;$ $\dfrac{b}{a} = k$ $\implies$ $b = ak$ $\;\;\; \cdots \; (1)$
$\dfrac{c}{b} = k$ $\implies$ $c = bk = ak^2$ $\;\;\;$ [by equation $(1)$] $\;\;\; \cdots \; (2)$
$\dfrac{d}{c} = k$ $\implies$ $d = ck = ak^3$ $\;\;\;$ [by equation $(2)$] $\;\;\; \cdots \; (3)$
Now, $\;$ $\left(a - c\right)^2 + \left(b - c\right)^2 + \left(b - d\right)^2 - \left(a - d\right)^2$
$= \left(a - ak^2\right)^2 + \left(ak - ak^2\right)^2 + \left(ak - ak^3\right)^2 - \left(a - ak^3\right)^2$
$\hspace{4cm}$ [by equations $(1)$, $(2)$ and $(3)$]
$= a^2 \left(1 - k^2\right)^2 + a^2 k^2 \left(1 - k\right)^2 + a^2 k^2 \left(1 - k^2\right)^2 - a^2 \left(1 - k^3\right)^2$
$= a^2 \left[\left(1 - k^2\right)^2 - \left(1 - k^3\right)^2\right] + a^2 k^2 \left[\left(1 - k\right)^2 + \left(1 - k^2\right)^2\right]$
$= a^2 \left[\left(1 + k\right)^2 \left(1 - k\right)^2 - \left(1 - k\right)^2 \left(1 + k + k^2\right)^2\right]$
$\hspace{3cm}$ $+ a^2 k^2 \left[\left(1 - k\right)^2 + \left(1 + k\right)^2 \left(1 - k\right)^2\right]$
$= a^2 \left(1 - k\right)^2 \left[\left(1 + k\right)^2 - \left(1 + k + k^2\right)^2\right] + a^2 k^2 \left(1 - k\right)^2 \left[1 + \left(1 + k\right)^2\right]$
$= a^2 \left(1 - k\right)^2 \left(1 + k + 1 + k + k^2\right) \left(1 + k - 1 - k - k^2\right)$
$\hspace{4cm}$ $+ a^2 k^2 \left(1 - k\right)^2 \left(1 + 1 + 2k + k^2\right)$
$= a^2 \left(1 - k\right)^2 \left(2 + 2k + k^2\right) \left(-k^2\right) + a^2 k^2 \left(1 - k\right)^2 \left(2 + 2k + k^2\right)$
$= 0$