Solve for $x$
$$2L_2 x^3 =2L_{1}(m-x)^3$$

Question

Solve for $x$
$$2L_2 x^3 =2L_{1}(m-x)^3$$

parthkohli · Answer

What do $\rm L_1$ and $\rm L_2$ stand for?

parthkohli · Answer

Dividing both sides by 2,$$\rm L_2 x^3 =L_1(m  - x)^3$$

anonymous · Answer

You don't need to know what they are..this is a minor problem from a larger problem..you can just treat it as some random variable..

I just need $x$

parthkohli · Answer

Okey dokey.

anonymous · Answer

And you need to remind me how to solve it..I don't seem to get it; to continue solving this problem

parthkohli · Answer

$$\rm L_2 x^3=L_1(m^3 -3m^2x+3mx^2 - x^3)$$I'm just gonna say that L1 = m and L2 = n

parthkohli · Answer

no. L1 = a and L2 = b.

parthkohli · Answer

$$\rm bx^3 = a(m^3 - 3m^2x + 3mx^2 - x^3)$$

anonymous · Answer

Oh no. I want a simple way please!

anonymous · Answer

It's some algebra law/ rulles that i forgot

parthkohli · Answer

Simple way? http://wolframalpha.com Enjoy!

anonymous · Answer

Lol, doesn't work. That is why I'm here :P

parthkohli · Answer

Step by step solution. $\rm  \checkmark $

anonymous · Answer

lol, does not work..it doesn't compute

parthkohli · Answer

The way I was doing it is actually the simplest. lol

anonymous · Answer

Not really :)$$L_2 x^3 =L_{1}(m-x)^3 =>\sqrt[3]{L_{2}}x=\sqrt[3]{L_{1}}  (m-x)$$
But i dont get what rules it comes froms?

parthkohli · Answer

Oh, it's just finding the cube root of both sides.

parthkohli · Answer

HOW DIDN"T THAT COME TO MY MIND? lol

anonymous · Answer

lol, i cant remmember how to do this

anonymous · Answer

so just assume that you can cube root both sides?

parthkohli · Answer

@Omniscience Are you seriously serious? You don't know how to cube root both sides

parthkohli · Answer

You can always cube root both sides. -.-

parthkohli · Answer

Whatever you do, do the same to both sides.

anonymous · Answer

lol..i forgot..but i thought it was only one side?

parthkohli · Answer

Both sides. BOTH SIDES ALWAYS

parthkohli · Answer

You gotta maintain the balance.

anonymous · Answer

$$L_2 x^3 =L_{1}(m-x)^3 => L_2 = \sqrt[3]{L_1(m-x)^2}$$
i thought you have to do it like this..

parthkohli · Answer

But also remember,$$\sqrt[3]{ab} = \sqrt[3]{a} \times \sqrt[3]{b}$$

anonymous · Answer

I know that..but why do you have to cube roots both sides? it just not make sense lol

parthkohli · Answer

BECAUSE BOTH SIDES ARE EQUAL.

anonymous · Answer

lol..can you do the for sqrt as well?

parthkohli · Answer

Yeah, all radicals.

anonymous · Answer

srsly!!!

parthkohli · Answer

Yeah. -.-

anonymous · Answer

I feel like an idiot lol

anonymous · Answer

:'(