If Rolle's theorem holds for the function f(x) = x ^ 3 - a x ^ 2 +...

$⇒ 1 - a + b - 4 = 8 - 4a + 2b - 4$

$⇒ 3a - b = 7 \ldots \left(1\right)$

Also ${f}^{'}\left(\dfrac{4}{3}\right) = 0$ (given)

$⇒ {\left(3{x}^{2} - 2ax + b\right)}_{x = \dfrac{4}{3}} = 0$

$⇒ \dfrac{16}{3} - \dfrac{8a}{3} + b = 0$

$⇒ 8a - 3b - 16 = 0 \ldots \left(2\right)$

Solving \left(1\right)$and \left(2\right)$

$a = 5, b = 8$