Explanation
$$f(x)=(\frac{\sqrt[3]{x^2-5x+6}}{x^2-x-6})$$
$$Cube\>root\>expression\>is\>fine\>for\>any\>value\>of\>the\>term\>inside\>the\>root.$$
$$But\>the\>denomiantor\>must\>not\>be\>equal\>to\>zero.$$
$$x^2-x-6\neq0$$
$$(x-3)(x+2)\neq0$$
$$x\neq3\>or\>x\neq-2$$
$$\therefore$$ domain: R- {-2,3}
Please disable the adBlock and continue. Thank you.