Solve the cubic polynomial $a^3 - 10a + 5 = 0$

Question

I am not able to understand how to solve a cubic polynomial. I have tried hit and trial for finding the first root but in vain. How do I solve for $a$ in $$ a^3 - 10a + 5 = 0\ ? $$

Answer 1

This equation has no integer solutions. You can approximate them numerically. For instance with Newton's method, or calculate them with the Cardano formula. https://www.encyclopediaofmath.org/index.php/Cardano_formula

Answer 2

Let $a=2\sqrt{\frac{10}{3}}\cos\phi.$

Thus, $$\frac{80\sqrt{10}\cos^3\phi}{3\sqrt3}-\frac{20\sqrt{10}\cos\phi}{\sqrt{3}}+5=0$$ or $$4\cos^3\phi-3\cos\phi=-\frac{3\sqrt3}{4\sqrt{10}}$$ or $$\cos3\phi=-\frac{3\sqrt3}{4\sqrt{10}}$$ and from here we can get three real roots.

Answer 3

$$a^3-10a+5=0$$ Set $a=u+v$ $$(u+v)^3-10(u+v)+5=0$$ Set $uv=\dfrac{10}{3}$ $$(u+v)^3-3uv(u+v)+5=0$$ $$u^3+3u^2v+3uv^2+v^3-3u^2v-3uv^2+5=0$$ $$u^3+v^3=-5;\;(uv)^3=u^3v^3=\frac{1000}{27}$$ Now consider an auxiliary unknown $z$

As we know a quadratic equation can be written as

$z^2-(z_1+z_2)z+z_1z_2=0$

We set $z_1=u^3;\;z_2=v^3$ so we get

$z^2-(u^3+v^3)z+u^3v^3=0$

and finally

$z^2+5z+\dfrac{1000}{27}=0$

which gives

$$z_1=u^3=\frac{5}{18} \left(-9-i \sqrt{399}\right);\;z_2=v^3=\frac{5}{18} \left(-9+i \sqrt{399}\right)$$

Now we have to compute the three cubic roots of the previous numbers, which is quite tedious. Anyway we have

$$u_1=\frac{i \sqrt[3]{\frac{5}{2} \left(\sqrt{399}-9 i\right)}}{3^{2/3}};\;u_2=-\frac{\sqrt[6]{-1} \sqrt[3]{\frac{5}{2} \left(\sqrt{399}-9 i\right)}}{3^{2/3}};\;u_3=-\frac{\sqrt[6]{-1}\sqrt[3]{\frac{5}{2} \left(\sqrt{399}-9 i\right)}}{3^{2/3}}$$ $$v_1=-\frac{i \sqrt[3]{\frac{5}{2} \left(\sqrt{399}+9 i\right)}}{3^{2/3}};\;v_2=\frac{\sqrt[6]{-1} \sqrt[3]{\frac{5}{2} \left(\sqrt{399}+9 i\right)}}{3^{2/3}};\;v_3=\frac{\sqrt[6]{-1} \sqrt[3]{\frac{5}{2} \left(\sqrt{399}+9 i\right)}}{3^{2/3}}$$ The three roots of the given equation are

$$a_1=u_1+v_1=\frac{i \sqrt[3]{\frac{5}{2}} \left(\sqrt[3]{\sqrt{399}-9 i}-\sqrt[3]{\sqrt{399}+9 i}\right)}{3^{2/3}}\approx 0.513544$$ $$a_2=u_2+v_2=\frac{\sqrt[6]{-1} \sqrt[3]{\frac{5}{2}} \left((-1)^{2/3} \sqrt[3]{\sqrt{399}+9 i}-\sqrt[3]{\sqrt{399}-9 i}\right)}{3^{2/3}}\approx -3.38762$$

$$a_3=u_3+v_3=\frac{\sqrt[6]{-1} \sqrt[3]{\frac{5}{2}} \left(\sqrt[3]{\sqrt{399}+9 i}-(-1)^{2/3} \sqrt[3]{\sqrt{399}-9 i}\right)}{3^{2/3}}\approx 2.87408$$

Now it is probably clearer the reason why numerical solutions are much more practical and fast and easier to find.

For instance this simple table, where $P(a)=a^3-10a+5$

$ \begin{array}{r|r} a & p(a)\\ \hline -4 & -19 \\ -3 & 8 \\ -2 & 17 \\ -1 & 14 \\ 0 & 5 \\ 1 & -4 \\ 2 & -7 \\ 3 & 2 \\ \end{array} $

allows us to say that a solution must be between $-4$ and $-3$, another one between $0$ and $1$ and the third between $2$ and $3$, because the value of the polynomial changes the sign and therefore in that interval it must take the value zero.

Now we can apply Newton's method starting with $a_0=2.5$

define recursively $a_{n+1}=a_n-\dfrac{p(a_n)}{p'(a_n)}$

where $p'(a)=3a^2-10$ is the derivative

we have

$ \begin{array}{r|r} n & a_n\\ \hline 0 & 2.5 \\ 1 & 3. \\ 2 & 2.88235 \\ 3 & 2.87412 \\ 4 & 2.87408 \\ 5 & 2.87408 \\ \end{array} $

so $a_3\approx 2.87408$

if we start from $a_0=0.5$

$ \begin{array}{r|r} n & a_n\\ \hline 0 & 0.5 \\ 1 & 0.513514 \\ 2 & 0.513544 \\ \end{array} $

so $a_2\approx 0.513544$

finally if we start from $a_0=-3.5$

$ \begin{array}{r|r} n & a_n \\ \hline 0 & -3.5 \\ 1 & -3.39252 \\ 2 & -3.38763 \\ 3 & -3.38762 \\ 4 & -3.38762 \\ \end{array} $

$a_1\approx -3.38762$

I hope this helps