Question

How do you solve the equation ${x^3} - 3x + 1 = 0$ ?

Answer 1

Hint: We will substitute some value of x and try to simplify the equation. We will choose the value of x such that it helps us to reduce the equation. We should be familiar with trigonometric formula like $\cos \dfrac{{2\pi }}{3} = - \dfrac{1}{2}$ , $\cos 3a = 4{\cos ^3}a - 3\cos a$ which are very helpful in solving these types of questions.

Complete step-by-step answer:
We have given equation ${x^3} - 3x + 1 = 0$
We will substitute some value for x and simplify the equation
We substitute $x = k\cos a$
So, the equation becomes
$ \Rightarrow {\left( {k\cos a} \right)^3} - 3\left( {k\cos a} \right) + 1 = 0$
We have taken k common from first and second term
$ \Rightarrow {k^3}{\cos ^3}a - 3k\cos a + 1 = 0$
$ \Rightarrow k\left( {{k^2}{{\cos }^3}a - 3\cos a} \right) + 1 = 0$
We will substitute k=2
We have taken the value of k=2 because we know that $\cos 3a = 4{\cos ^3}a - 3\cos a$ and it will help us to reduce the equation.
$ \Rightarrow 2\left( {4{{\cos }^3}a - 3\cos a} \right) + 1 = 0$
$ \Rightarrow 2\cos 3a + 1 = 0$
So, $\cos 3a = - \dfrac{1}{2}$
We also know that $\cos \dfrac{{2\pi }}{3} = - \dfrac{1}{2}$$ \Rightarrow a = \pm \left( {\dfrac{{2 + 6k}}{9}} \right)\pi $
$ \Rightarrow 3a = \pm {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) + 2k\pi $
$ \Rightarrow 3a = \pm \dfrac{{2\pi }}{3} + 2k\pi $
$ \Rightarrow a = \pm \left( {\dfrac{{2 + 6k}}{9}} \right)\pi $
Hence, the solution of the equation ${x^3} - 3x + 1 = 0$ is $ \pm \left( {\dfrac{{2 + 6k}}{9}} \right)\pi $ for any integer of k.

Note: It's important to note that a polynomial equation is a set of variables and their coefficients; the preceding equation is a third-degree polynomial equation. We have to find the value of x which satisfies the equation and since we have a third degree equation, we can have a maximum root is three.