Simplify 3 sqrt(12) / (4 * cube root of 54), leaving no radical in the denominator.

Step-by-step solution

1. 1

   Simplify each radical before doing anything else. Look for the largest perfect square inside the square root and the largest perfect cube inside the cube root:

   $\sqrt{12}=\sqrt{4\cdot 3}=2\sqrt3,\qquad \sqrt[3]{54}=\sqrt[3]{27\cdot 2}=3\sqrt[3]{2}$

   Matching the index to the power is the key habit: squares come out of a square root, cubes out of a cube root.

2. 2

   Substitute and cancel the rational coefficients. The fraction becomes

   $\frac{3\cdot 2\sqrt3}{4\cdot 3\sqrt[3]{2}}=\frac{6\sqrt3}{12\sqrt[3]{2}}=\frac{\sqrt3}{2\sqrt[3]{2}}$

   Only the numbers cancel here; $\sqrt3$ and $\sqrt[3]{2}$ have different indices, so they cannot be combined.

3. 3

   Pick the right rationalising factor for a cube root. For a square root you multiply by the same root; for $\sqrt[3]{2}$ that is not enough, because $\sqrt[3]{2}\cdot\sqrt[3]{2}=\sqrt[3]{4}$ is still irrational. You need a total of three factors of 2 under the radical, so multiply top and bottom by $\sqrt[3]{4}$:

   $\frac{\sqrt3}{2\sqrt[3]{2}}\cdot\frac{\sqrt[3]{4}}{\sqrt[3]{4}}=\frac{\sqrt3\,\sqrt[3]{4}}{2\sqrt[3]{8}}$

4. 4

   Finish the denominator. Since $\sqrt[3]{8}=2$,

   $\frac{\sqrt3\,\sqrt[3]{4}}{2\cdot 2}=\frac{\sqrt3\,\sqrt[3]{4}}{4}$

   The denominator is now the rational number $4$, as required.

5. 5

   Check the decimal value. The original is $\dfrac{3\cdot 3.4641}{4\cdot 3.7798}=0.68736$, and $\dfrac{1.7321\cdot 1.5874}{4}=0.68736$. The two agree, so the simplification is correct.