If α , β , γ are cube roots of -8 , then what is (α^2 p^2+β^2...

Since $\alpha, \beta, \gamma$ are the cube roots of $-8$, they are the roots of the equation $z^3 + 8 = 0$.

Thus, we have $\alpha^3 = \beta^3 = \gamma^3 = -8$.

The product of the roots of the cubic equation $z^3 + 8 = 0$ is given by $\alpha \beta \gamma = -8$.

Using these properties, we can find relations between the roots:

$\alpha^3 = \alpha \beta \gamma \Rightarrow \alpha^2 = \beta \gamma$

$\beta^3 = \alpha \beta \gamma \Rightarrow \beta^2 = \alpha \gamma$

$\gamma^3 = \alpha \beta \gamma \Rightarrow \gamma^2 = \alpha \beta$

Let the given expression be $E = \dfrac{\alpha^2 p^2+\beta^2 q^2+\gamma^2 r^2}{\beta^2 p^2+\gamma^2 q^2+\alpha^2 r^2}$.

Consider multiplying the denominator by the fraction $\dfrac{\gamma}{\beta}$:

$\dfrac{\gamma}{\beta} (\beta^2 p^2 + \gamma^2 q^2 + \alpha^2 r^2) = \beta \gamma p^2 + \dfrac{\gamma^3}{\beta} q^2 + \dfrac{\alpha^2 \gamma}{\beta} r^2$

Now, substitute the derived relations into each term of this expansion:

For the first term, substitute $\beta \gamma = \alpha^2$ to get $\alpha^2 p^2$.

For the second term, since $\gamma^3 = \beta^3 = -8$, we have $\dfrac{\gamma^3}{\beta} = \dfrac{\beta^3}{\beta} = \beta^2$, which gives $\beta^2 q^2$.

For the third term, substitute $\alpha^2 = \beta \gamma$ to get $\dfrac{(\beta \gamma) \gamma}{\beta} r^2 = \gamma^2 r^2$.

Adding these terms together, we get:

$\dfrac{\gamma}{\beta} (\beta^2 p^2 + \gamma^2 q^2 + \alpha^2 r^2) = \alpha^2 p^2 + \beta^2 q^2 + \gamma^2 r^2$

This shows that the numerator is exactly $\dfrac{\gamma}{\beta}$ times the denominator.

Therefore, the value of the given expression is $\dfrac{\gamma}{\beta}$.

Answer: $\dfrac{\gamma}{\beta}$