How to get the simplest form of this radical expression: $3\sqrt[3]{2a} - 6\sqrt[3]{2a}$.

Question

How to get the simplest form of this radical expression: $$3\sqrt[3]{2a} - 6\sqrt[3]{2a}$$

Here is my work: $$3\sqrt[3]{2a} - 6\sqrt[3]{2a}$$ Since the radicands are the same, we just add the coefficients. $$-3\sqrt[3]{2a} \sqrt[3]{2a}$$ Since everything is under the same index it becomes: $$-3\sqrt[3]{2} \sqrt[3]{a}$$

Did I do this correctly, if not can anyone tell me what I should do? Thanks :-).

Answer 1

Your middle step is incorrect, it should be $-3\sqrt[3]{2a}$ not $-3\sqrt[3]{2a}\sqrt[3]{2a}$. It should be $$3\sqrt[3]{2a} - 6\sqrt[3]{2a} = 3\times\sqrt[3]{2a} - 6\times\sqrt[3]{2a} = (3 - 6)\times\sqrt[3]{2a} = -3\times\sqrt[3]{2a} = -3\sqrt[3]{2a}.$$ I don't think $-3\sqrt[3]{2}\sqrt[3]{a}$ is any simpler than $-3\sqrt[3]{2a}$, but that's just my opinion.

Answer 2

We can factor, as an alternative, to get the same result:

$$3\sqrt[3]{2a} - 6\sqrt[3]{2a} = 3\sqrt[3]{2a}(1 - 2) = -3\sqrt[3]{2a}$$

(I don't see any need to write: $\;-3\sqrt[3]{2a} = -3\sqrt[3]{2} \sqrt[3]{a})$

Note that $-3\sqrt[3]{2a}\sqrt[3]{2a} = -3\sqrt[3]{4a^2} \neq -3\sqrt[3]{2a} $

Answer 3

$$3\sqrt[3]{2a}-6\sqrt[3]{2a}=\sqrt[3]{2a}(3-6)=\sqrt[3]{2a}(-3)= -3\sqrt[3]{2a}$$