Method 1

Given that $a,b,$ and $c$ are in arithmetic progression (A.P), so we have the relationship:

$b=\frac{a+c}{2}$

$\Rightarrow 2b=a+c$ ...(i)

We need to find the value of $a^3+c^3-8b^3$.

$a^3+c^3-8b^3=a^3+c^3+(-2b{)}^3$

Here $a+c+(-2b)=2b-2b=0$

$[a+c=2b,\text{Using (i)}]$

.

Therefore, $a^3+c^3+(-2b{)}^3=3ac(-2b)=-6abc$.

[Here we used the identity $x^3+y^3+z^3=3xyz$, if $x+y+z=0$]

Method 2

Given that $a,b,$ and $c$ are in arithmetic progression (A.P), so we have the relationship:

$b=\frac{a+c}{2}$

$\Rightarrow 2b=a+c$ ...(i)

By taking cube on both sides of $2b=a+c$, we get:

$(2b{)}^3=(a+c{)}^3$

$8b^3=a^3+c^3+3ac(a+c)$

$\Rightarrow a^3+c^3-8b^3=-3ac(a+c)$

$\Rightarrow a^3+c^3-8b^3=-3ac(2b)$.

$[a+c=2b,\text{Using (i)}]$

$\Rightarrow a^3+c^3-8b^3=-6abc$