A rationalising factor of $(\sqrt[3]{9} - \sqrt[3]{3} + 1)$ is

Concept & Formula To rationalise an expression containing cube roots, we use the standard algebraic identities for the sum or difference of cubes. $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$ We need to multiply the given irrational expression by a factor that transforms it into a rational integer (like $a^3 + b^3$). Step-by-Step Solution * **Given expression**: $\sqrt[3]{9} - \sqrt[3]{3} + 1$ * Rewrite the expression to identify the algebraic structure. Note that $9 = 3^2$. $$(\sqrt[3]{3})^2 - \sqrt[3]{3}(1) + (1)^2$$ * Let $a = \sqrt[3]{3}$ and $b = 1$. The expression exactly matches the quadratic factor of the sum of cubes formula: $$(a^2 - ab + b^2)$$ * To rationalize it (meaning, to get rid of the cube roots by turning the whole thing into $a^3 + b^3$), we must multiply it by the missing factor: $$(a + b)$$ * Substitute $a$ and $b$ back into the missing factor: $$(\sqrt[3]{3} + 1)$$ * If we multiply them, the result is $(\sqrt[3]{3})^3 + 1^3 = 3 + 1 = 4$, which is perfectly rational. Exam Strategy & Shortcut Recognize the pattern $a^2 - ab + b^2$. The presence of the mixed minus and plus signs ($+$, $-$, $+$) always pairs with the factor $(a + b)$ to create a sum of cubes ($a^3 + b^3$). Here, the base elements are clearly $\sqrt[3]{3}$ and $1$. Therefore, the rationalizing multiplier is simply their sum: $\sqrt[3]{3} + 1$. Common Pitfall Confusing the sum of cubes with the difference of cubes. If a student sees the minus sign and instinctively chooses $(\sqrt[3]{3} - 1)$, the resulting multiplication would not clear the radicals properly, as the signs in the trinomial would be mismatched. Final Answer **Therefore, the correct answer is $\sqrt[3]{3} + 1$.**