Can one of you help me w/ the following equation?

Can one of you help me w/ the following equation?

p^3 - 3p = 4

I'm trying to get back into maths and am stumped.

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wolframalpha.com/input/?i=p^3 - 3p = 4
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by the rational root test (en.wikipedia.org/wiki/Rational_root_theorem) there's no rational roots since -4,-2,-1,1,2,4 aren't roots, so all the roots are irrational

wolframalpha.com/input/?i=p^3 - 3p = 4
p=2.1958...
p=-1.097...+0.785..i
p=-1.097...-0.785..i

so p^3-3p=4 can be written
(p-2.1958...)(p-1.097...+0.785..i)(p-1.097...-0.785..i)=0

So, you want a step-by-step for [math]p^3 - 3 p = 4[/math]

Answer: [math]p = 1/(2 + sqrt(3))^(1/3) + (2 + sqrt(3))^(1/3) or p = ((-1)^(2/3) - (-1)^(1/3) (2 + sqrt(3))^(2/3))/(2 + sqrt(3))^(1/3) or p = (-1)^(2/3) (2 + sqrt(3))^(1/3) - (sqrt(3) - 2)^(1/3)[/math]

!WORKING IT OUT FO YOSELF!

First Hint: Move everything to the left hand side by subtracting 4 from both sides.
p^3 - 3 p - 4 = 0

Hint: Perform the substitution p = x + λ/x.
Change coordinates by substituting p = x + λ/x, where λ is a constant value that will be determined later: -4 - 3 (x + λ/x) + (x + λ/x)^3 = 0

Hint: Transform the rational equation into a polynomial equation.
Multiply both sides by x^3 and collect in terms of x: x^6 + x^4 (3 λ - 3) - 4 x^3 + x^2 (3 λ^2 - 3 λ) + λ^3 = 0

Hint: Find an appropriate value for λ in order to make the coefficients of x^2 and x^4 both zero.
Substitute λ = 1 and then y = x^3, yielding a quadratic equation in the variable y: y^2 - 4 y + 1 = 0

Hint: Solve for y by finding the positive solution to the quadratic equation: y = 2 + sqrt(3)

Hint: Perform back substitution on y = 2 + sqrt(3), so you gotta substitute back for y = x^3:
x^3 = 2 + sqrt(3)

Hint: Take the cube root of both sides. Taking cube roots gives (2 + sqrt(3))^(1/3) times the third roots of unity:
x = (2 + sqrt(3))^(1/3) or x = -(-2 - sqrt(3))^(1/3) or x = (-1)^(2/3) (2 + sqrt(3))^(1/3)

Hint: Perform back substitution with p = x + 1/x, which means substitute each value of x into p = x + 1/x:
p = 1/(2 + sqrt(3))^(1/3) + (2 + sqrt(3))^(1/3) or p = (-1)^(2/3)/(2 + sqrt(3))^(1/3) - (-2 - sqrt(3))^(1/3) or p = (-1)^(2/3) (2 + sqrt(3))^(1/3) - (-1/(2 + sqrt(3)))^(1/3)

Final Hint: Bring each solution to a common denominator and simplify.

Bleh, how ugly. Forgive me.

Answer: [math]p=\frac{1}{3} \sqrt[3]{54-27 \sqrt{3}}+\sqrt[3]{2+\sqrt{3}}[/math]

Are you sure it's not p^2 - 3p = 4?
If it is, notice that you can re-arrange it to become, p^2 - 3p - 4 = 0, and by inspection of the quadratic we have that -4 are numbers so that -4 + 1 = -3, and -4 * -1 = -4, meaning we can write our quadratic as (p-4)(p+1) = 0, and so the solutions are p = 4, and p = -1.

If its actually a cubic, then note that it has no rational solution, and even by transforming it by putting p = 2*cos(x) into it, we obtain cos(3x) = 2, which has no real solutions.

>If its actually a cubic, then note that it has no rational solution, and even by transforming it by putting p = 2*cos(x) into it, we obtain cos(3x) = 2, which has no real solutions.
>cubic
>no real solutions

im sure you understood the slight error in the context of what i was saying, that cos(3x) = 2 has no real solutions

>that cos(3x) = 2 has no real solutions
Why is that relevant?

>If its actually a cubic

Evaluate your prime statement before depth traversal.

>Evaluate your prime statement before depth traversal.
What's the relevance of cos(3x)=2 having no real solution?