True or false or it depends?

If $a\left(\dfrac{1}{b} + \dfrac{1}{c}\right) , b\left(\dfrac{1}{c} + \dfrac{1}{a}\right) , c\left(\dfrac{1}{a} + \dfrac{1}{b}\right)$ is in AP,
$a\left(\dfrac{1}{b} + \dfrac{1}{c}\right) + 1 , b\left(\dfrac{1}{c} + \dfrac{1}{a}\right) + 1 , c\left(\dfrac{1}{a} + \dfrac{1}{b}\right) + 1$ is in A.P.
i.e. $a\left(\dfrac{1}{b} + \dfrac{1}{c}\right) + \dfrac{a}{a} , b\left(\dfrac{1}{c} + \dfrac{1}{a}\right) + \dfrac{b}{b} , c\left(\dfrac{1}{a} + \dfrac{1}{b}\right) + \dfrac{c}{c}$ is in A.P.
i.e. $a\left(\dfrac{1}{b} + \dfrac{1}{c} + \dfrac{1}{a}\right) , b\left(\dfrac{1}{c} + \dfrac{1}{a} + \dfrac{1}{b}\right) , c\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right)$ is in A.P.
i.e. $a,b,c$ are in A.P.

So, $2b = a + c\\ \therefore b = \dfrac{a + c}{2} \longrightarrow \boxed{1}$

Now take the second equation:-
$a^3 + c^3 + 6abc = 8b^3$

Substituting $\boxed{1}$ in the above equation, we get:-
$a^3 + c^3 + 6a × \dfrac{a + c}{2} × c = 8{\left(\dfrac{a + c}{2}\right)}^3\\ (a^3 + c^3 + 3ac(a + c) = 8 \dfrac{{(a + c)}^3}{8}\\ {(a + c)}^3 = {(a + c)}^3$ .

Since the equation satisfies, it is proved that $a^3 + c^3 + 6abc$ is equal to $8b^3$

So, the answer is $\color{#69047E}{\boxed{\text{True}}}$ .