1. ## Changing the subject

(x+a)^3 / x = c

Express x in terms of a and c .

How do you do it?

2. Originally Posted by hitmen
(x+a)^3 / x = c

Express x in terms of a and c .

How do you do it?
With an enormous amount of difficulty. Solving a general cubic equation is a serious pain. Are you sure that exponent isn't a 2, not a 3?

-Dan

3. Well, this isnt a textbook question. I cant remember where I saw this. If it is difficult can you just give me the appropriate steps to be taken. Thanks. I should be able to work it out myself.

4. Hi !

It's not easy for general formulae...
If you know one of the roots, then you will easily be able to find the others (polynomial factorisation).

Now, I know there is a general formula for $x^3+px+q=0$

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To refer to your exercise, you're looking for :

$(x+a)^3=cx \Longleftrightarrow (x+a)^3-cx=0$
you can change x+a for y :

$y^3-cy+ca=0$

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If $4p^3+27q^2 \ge 0$, then there is a real solution (& 2 complex ones) :

$x=\sqrt[3]{-\frac q2+\sqrt{\left(\frac q2\right)^2+\left(\frac p3\right)^3}}+\sqrt[3]{-\frac q2-\sqrt{\left(\frac q2\right)^2+\left(\frac p3\right)^3}}$

Unfortunately, I can't remember who discovered it (he was an Italian guy, in the XVI th century if my memory is correct)... I had to give back the book to the library.

Edit : his name is Bombelli.

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I've just seen you can use the Cardan method or trigonometric method... Google for it, I don't have enough knowledge to talk about it.

5. See Cardano's method.

-Dan

6. ## solving cubic equation

even i do substitution, what is next?

y=x+a
y^3 = cy-ca
y^3 -cy +ca = 0

how to factorise or solve?

BTW, how do you enter square roots, power and other symbols?

7. now this is the only thing I dont understand.

Setting 3uv + p = 0

Why?

8. Originally Posted by hitmen
now this is the only thing I dont understand.

Setting 3uv + p = 0

Why?
If you look at the general form as you go along you will note that the derivation adds one extra variable to the solution, giving us the freedom to choose a single condition that is convenient for us. This is the condition that we choose. Once this substitution is made the equation falls into something like the form
$p^3 + \frac{n}{p^3} = m$

$(p^3)^2 - mp^3 + n = 0$

This is a "tri-quadratic" equation, I guess you'd call it. We can easily solve this for p by using the quadratic formula to solve for $p^3$ then simply taking the cube root to find p. Then you can back substitute to get your answer in terms of the original coefficients.

You can get a tutorial on LaTeX here, in the first few posts of the sticky thread at the top of the list.

-Dan