solution:
f(x) has discontinuity at x=3
if it is continuous at x=3
then the continuous on (-\infty ,\infty )
we have,
f(x)= \left \{cx^{2}+8 ; if x< 3,x^{3}-cx ;if x\geq3
\lim_{x\rightarrow 3^-}f(x)=\lim_{x\rightarrow3^+}f(x)
\lim_{x\rightarrow3^-}(x^{2}+8x)=\lim_{x\rightarrow3^+}(x^{3}-cx)
(9c+24)= (227-3c)
9c+24=27-3c
9c+3c=27-24
12c=3
c=\frac{1}{4}
