# Simplify the odd root: `root(3)(24a^4)` ``

baxthum8 | High School Teacher | (Level 3) Associate Educator

Posted on

In order to simplify `root(3)(24a^4)` we must first factorize both 24 into perfect cubes.

Meaning: perfect cubes include:  `2^3 = 8; 3^3 = 27; 4^3 = 64, etc`

Since 24 = 8 x 3, we can rewrite 24 as `2^3* 3`  and we can rewrite `a^4` as `a^3* a` .

we still also have 3 and a left over from our factorization.

Now I can simplify: `root(3)(2^3*a^3)`

Now we will have:  `2aroot(3)(3a)`

durbanville | High School Teacher | (Level 2) Educator Emeritus

Posted on

To simplify `root(3)(24a^4)`

first simplify the `24a^4`

`therefore root(3)(24a^4)= root(3)(3 times 2^3 times a^4)`

As we have a cube root (hence an "odd" root), we can simplify as follows:

`therefore = root(3)(3)root(3)(2^3)root(3)(a timesa timesa times a)`

Note how we can cancel the root 3 against a cube in `root (3)(2^3)`  and with the a-s:

`therefore = 2aroot(3)(3a)`

Ans:

`root(3)(24a^4) =`  `2a root(3)(3a)`

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