forgive me but i dont know how to do this in Latex

cubed root(-81xcubed) - 2x cubed root(3) + 5x cubed root(24)

i dont need to solve i just need to simplify.
I think i have some idea like for cubed root 81 do i find a smaller number cubed to take out? what?

2. Originally Posted by MistaMista
forgive me but i dont know how to do this in Latex

cubed root(-81xcubed) - 2x cubed root(3) + 5x cubed root(24)

i dont need to solve i just need to simplify.
I think i have some idea like for cubed root 81 do i find a smaller number cubed to take out? what?
That's ok, in LaTeX, this would be:

Code:
\sqrt[3]{-81x^3}-2x\sqrt[3]{3}+5x\sqrt[3]{24}
Be sure to put the code in the "math" tags

$\sqrt[3]{-81x^3}-2x\sqrt[3]{3}+5x\sqrt[3]{24}$

Simplifying, we get

$\sqrt[3]{(-3)^3x^3}-2\sqrt[3]{3}x+5x\sqrt[3]{2^3\times 3}$

$\implies -3x-2\sqrt[3]{3}x+5x(2\sqrt[3]{3})$

$\implies (-3-2\sqrt[3]{3}+10\sqrt[3]{3})x$

$\implies\color{blue}\boxed{(-3+8\sqrt[3]{3})x}$

Does this makes sense?

--Chris

3. ## no sense

Can you explain how you did this b/c i am wondering how your moving from step to step

4. Hello, MistaMista!

$\sqrt[3]{-81x^3} - 2x\sqrt[3]{3} + 5x\sqrt[3]{24}$

I think i have some idea like for $\sqrt[3]{81}$
Do i find a smaller number cubed to take out? . . . . Yes!

The first term is: . $\sqrt[3]{-81x^3} \:=\:\sqrt[3]{(-27)(3)x^3} \:=\:\sqrt[3]{-27}\!\cdot\!\sqrt[3]{3}\!\cdot\!\sqrt[3]{x^3} \;=\;-3x\sqrt[3]{3}$

The last term is: . $5x\sqrt[3]{24} \;=\;5x\sqrt[3]{(8)(3)} \;=\;5x\sqrt[3]{8}\!\cdot\!\sqrt[3]{3}\;=\;5x\!\cdot\!2\sqrt[3]{3} \;=\;10x\sqrt[3]{3}$

The problem becomes: . $-3x\sqrt[3]{3} - 2x\sqrt[3]{3} + 10x\sqrt[3]{3}$

They are all "similar" terms: . $(-3-2+10)x\sqrt[3]{3} \;\;=\;\;\boxed{5x\sqrt[3]{3}}$

5. ## thanks

i think its just going to take me some time to get it.

6. Originally Posted by Soroban
Hello, MistaMista!

The first term is: . $\sqrt[3]{-81x^3} \:=\:\sqrt[3]{(-27)(3)x^3} \:=\:\sqrt[3]{-27}\!\cdot\!\sqrt[3]{3}\!\cdot\!\sqrt[3]{x^3} \;=\;-3x\sqrt[3]{3}$

The last term is: . $5x\sqrt[3]{24} \;=\;5x\sqrt[3]{(8)(3)} \;=\;5x\sqrt[3]{8}\!\cdot\!\sqrt[3]{3}\;=\;5x\!\cdot\!2\sqrt[3]{3} \;=\;10x\sqrt[3]{3}$

The problem becomes: . $-3x\sqrt[3]{3} - 2x\sqrt[3]{3} + 10x\sqrt[3]{3}$

They are all "similar" terms: . $(-3-2+10)x\sqrt[3]{3} \;\;=\;\;\boxed{5x\sqrt[3]{3}}$

Forgot that $3^4=81, \ not \ 3^3=81$...

7. ## i get it nows

ok i understand. i was looking at it in the wrong way. Thanks a lot