Rolles's theorem holds for the function $x^3+\mathrm{b} x^2+\mathrm{c} x, 1 \leq x \leq 2$ at the point $\frac{4}{3}$, the respective values of $b$ and $c$ are :

$-5,8$

$f(x)=x^3+b x^2+c x \quad 1 \leq x \leq 2$

So by Rolle's theorem

f(x) is continous on [1, 2]

differentiable on (1, 2)

Hence $f(1)=f(2)$

$\Rightarrow (1)^3+b(1)^2+c(1)=(2)^3+(2)^2 b+(2) c$

$1+b+c=8+4 b+2 c$

$(8-1)+(4 b-b)+(2 c-c)=0$

7 + 3b + c = 0        .....(1)

$f'(x)=\frac{d}{d x}\left(x^3+b x^2+c\right)=3 x^2+2 b x+c-2$

at $4 / 3 \quad f'(4 / 3)=0$ as por rolle's theorem

f'(4/3) = 0

substituting from eq (2)

$=3 \times\left(\frac{4}{3}\right)^2+2 \times \frac{4}{3} b+c=0$

$\Rightarrow 3 \times \frac{16}{9}+\frac{8}{3} b+c=0$

$=\frac{16}{3}+\frac{8}{3} b+c=0$

$16+8 b+3 c=0$            ........(3)

eq (1) × (2)   ---  eq (3)

$3 \times(7+3 b+c=0)$

$-(16+8 b+3 c=0)$

$\Rightarrow 21+a b+3 c=0$

$-16+8 b+3 c=0 $

$5+b=0 \Rightarrow b =-5$

Sunstituting b = -5 in eq (1) → we get

$7 \times 3(-5)+c=0$

$7-15+c=0 \Rightarrow-8+c=0 \Rightarrow c=8$

b = -5, c = 8