The mechanical power P delivered to a particle depends on time t...

According to the principle of homogeneity of dimensions, quantities added or subtracted must have the same dimensions.

From the numerator $(\alpha - t^2)$, we have:
$[\alpha] = [t^2] = [\mathrm{T}^2]$

The dimensional formula for power $P$ is $[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-3}]$.

Substituting the dimensions into the given equation $P = \dfrac{\alpha - t^2}{\beta x}$:
$[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-3}] = \dfrac{[\mathrm{T}^2]}{[\beta] [\mathrm{L}]}$

Rearranging for $[\beta]$:
$[\beta] = \dfrac{[\mathrm{T}^2]}{[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-3}] [\mathrm{L}]} = \dfrac{[\mathrm{T}^2]}{[\mathrm{M} \mathrm{L}^3 \mathrm{~T}^{-3}]} = [\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^5]$

The dimensions of the product $(\alpha \times \beta)$ are:
$[\alpha \times \beta] = [\mathrm{T}^2] \times [\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^5] = [\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^7]$

Answer: $[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^7]$