Correct option is (C) \(\sqrt[3]4+1\)

A rationalising factor of a given irrational number is an irrational number whose multiple with given irrational number gives a rational number.

(A) \((16^\frac13-4^\frac13+1)(4^\frac13+2)\)

\(=64^\frac13-16^\frac13+4^\frac13+2.16^\frac13-2.4^\frac13+2\)

\(=(4^3)^\frac13+16^\frac13-4^\frac13+2\)

\(=4+2+16^\frac13-4^\frac13\)

\(=6+16^\frac13-4^\frac13\) which is an irrational number.

Thus, \((4^\frac13+2)\) is not a rationalising factor of \((16^\frac13-4^\frac13+1).\)

(B) \((16^\frac13-4^\frac13+1)(4^\frac13-2)\)

\(=64^\frac13-16^\frac13+4^\frac13-2.16^\frac13+2.4^\frac13-2\)

\(=(4^3)^\frac13-3.16^\frac13+3.4^\frac13-2\)

\(=4-2-3.16^\frac13+3.4^\frac13\)

\(=2-3.16^\frac13+3.4^\frac13\) which is an irrational number.

Thus, \((4^\frac13-2)\) is not a rationalising factor of \((16^\frac13-4^\frac13+1).\)

(C) \((16^\frac13-4^\frac13+1)(4^\frac13+1)\)

\(=64^\frac13-16^\frac13+4^\frac13+16^\frac13-4^\frac13+1\)

\(=(4^3)^\frac13+1\)

= 4+1 = 5 which is a rational number.

Thus, \((4^\frac13+1)\) is a rationalising factor of \((16^\frac13-4^\frac13+1).\)

(D) \((16^\frac13-4^\frac13+1)(4^\frac13-1)\)

\(=64^\frac13-16^\frac13+4^\frac13-16^\frac13+4^\frac13-1\)

\(=(4^3)^\frac13-2.16^\frac13+2.4^\frac13-1\)

\(=4-1-2.16^\frac13+2.4^\frac13\)

\(=3-2.16^\frac13+2.4^\frac13\) which is an irrational number.

Thus, \((4^\frac13-1)\) is not a rationalising factor of \((16^\frac13-4^\frac13+1).\)

Correct option is (C) \(^3\sqrt{4}\) + 1