If $x^m + y^m = k, (m \neq 1)$ and $y'' = \dfrac{ax^b}{y^c}$ such that $a + b + c = 0$, then the value of k is...

Differentiating $x^m + y^m = k$ with respect to $x$:

$m x^{m-1} + m y^{m-1} y' = 0$

$y' = -x^{m-1} y^{1-m}$

Differentiating again with respect to $x$:

$y'' = -(m-1) x^{m-2} y^{1-m} - x^{m-1} (1-m) y^{-m} y'$

Substituting $y' = -x^{m-1} y^{1-m}$:

$y'' = (1-m) x^{m-2} y^{1-m} - (1-m) x^{m-1} y^{-m} (-x^{m-1} y^{1-m})$

$y'' = (1-m) x^{m-2} y^{1-m} + (1-m) x^{2m-2} y^{1-2m}$

Factoring out $(1-m) x^{m-2} y^{1-2m}$:

$y'' = (1-m) x^{m-2} y^{1-2m} (y^m + x^m)$

Since $x^m + y^m = k$, we get:

$y'' = \dfrac{k(1-m) x^{m-2}}{y^{2m-1}}$

Comparing this with $y'' = \dfrac{ax^b}{y^c}$, we obtain:

$a = k(1-m)$

$b = m-2$

$c = 2m-1$

Given $a + b + c = 0$, we have:

$k(1-m) + (m-2) + (2m-1) = 0$

$k(1-m) + 3m - 3 = 0$

$k(1-m) - 3(1-m) = 0$

$(k - 3)(1 - m) = 0$

Since $m \neq 1$, it follows that $k = 3$.

Answer: $3$