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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 $$ a^3 - 10a + 5 = 0\ ? $$

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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.

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    $\begingroup$ Where the heck did $2\sqrt{\frac{10}{3}}\cos\phi$ come from? $\endgroup$ Oct 21, 2017 at 18:10
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    $\begingroup$ @user2357112 I want to get a possibility to use $\cos3\phi=4\cos^3\phi-3\cos\phi$. For which I wrote $x=k\cos\phi$ and solved an equation on $k$. $\endgroup$ Oct 21, 2017 at 18:14
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    $\begingroup$ That sounds like a general technique for solving any equation of the form $ax^3-bx+c=0$. It'd be nice to have the general case in the answer. $\endgroup$ Oct 21, 2017 at 18:20
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    $\begingroup$ @user2357112 Yes, you are right, it's a general way in the case, when the cubic equation has three real roots. For your equation we obtain: $x=k\cos\phi$, which gives $\frac{ak^3}{bk}=\frac{4}{3}$ and we can assume $k=\sqrt{\frac{4b}{3a}}$. Actually, this method does not help, when our equation has an unique real root. $\endgroup$ Oct 21, 2017 at 18:27
  • $\begingroup$ In the unique real root case, I suppose we get a value outside the $[-1, 1]$ range for $\cos^3\phi$. $\endgroup$ Oct 21, 2017 at 18:41
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$$a^3-10a+5=0$$ 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

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    $\begingroup$ Deserves more upvotes. $\endgroup$
    – Cornman
    Oct 21, 2017 at 18:13
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    $\begingroup$ The real and imaginary parts cube roots of arbitrary complex numbers in general can not be expressed as a finite sequence of arithmetic operations, square root extractions, and cube root extractions over the real numbers. Thus, even if a complex number is a solution to a cubic equations, and thus expressible as a finite sequence of arithmetic operations, square root extractions, and cube root extractions over the coefficients, its real and imaginary parts may not be so. $\endgroup$ Feb 25 at 21:32
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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

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We can easily check that $a^3-10a+5=0$ has no rational solution. For polynomials with leading coefficient 1 and constant coefficient $5$, only $-5, -1, 1,$ and $5$ are possible rational solutions, due to the Rational Root Theorem.

We apply the cubic formula. As there is no quadratic term, the formula is simplified.

https://math.stackexchange.com/a/4868629/928654

Let $i^2=-1$

and exactly one of the following be true

$w_1=w_2=1$

or $w_1=\frac{-1+i\sqrt3}{2}$,and$w_2=\frac{-1-i\sqrt3}{2}$

or $w_1=\frac{-1-i\sqrt3}{2}$, and $w_2=\frac{-1+i\sqrt3}{2}$

then

$$w_1\sqrt[3]{-2.5+\sqrt{6.25-\frac{1000}{27}}}+w_2\sqrt[3]{-2.5-\sqrt{6.25-\frac{1000}{27}}}=w_1\sqrt[3]{-2.5+\frac{i\sqrt3325}{6\sqrt3}}+w_2\sqrt[3]{-2.5-\frac{i\sqrt3325}{6\sqrt3}}$$

We have the cube root of a complex number, whose real and imaginary parts are not expressible by radicals.

Complex numbers have an absolute value. In this case, we have $\sqrt{2.5^2+(\frac{\sqrt3325}{6\sqrt3})^2}=\sqrt{6.25+\frac{3325}{108}}$. this is about $6.086$

The absolute value of the cube root of $2.5+\frac{i\sqrt3325}{6\sqrt3}$ is about $1.8257$

Complex numbers can be associated with a particular number of radians. Because the real part is negative, it is the inverse of the tangent function plus $\pi$ radians.

$\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi$ is about $4.2891$ radians

$\tan^{-1 rad}(\frac{-\sqrt3325}{15\sqrt3})$ is about $1.9941$ radians

Divide by $3$ and we get about $1.4327$ radians and $0.6647$ radians, respectively.

So we have

$$a=w_1\sqrt[6]{6.25+\frac{3325}{108}}(\cos(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3})+i\sin(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3}))+w_2\sqrt[6]{6.25+\frac{3325}{108}}(\cos(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3})-i\sin(\frac{\tan^{-1}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3}))$$

Note that the sums of complex conjugates are always real. For $w_1=w_2=1$, $a$ is real.

$$a_1=\sqrt[6]{6.25+\frac{3325}{108}}(\cos(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3})+i\sin(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3}))+\sqrt[6]{6.25+\frac{3325}{108}}(\cos(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3})-i\sin(\frac{\tan^{-1}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3}))=2\sqrt[6]{6.25+\frac{3325}{108}}(\cos(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3}))$$

this is about $0.5135$

But what of the other two solutions. Now we go with $w_1=\frac{-1+i\sqrt3}{2}$,and $w_2=\frac{-1-i\sqrt3}{2}$

$$a_2=(\frac{-1+i\sqrt3}{2})\sqrt[6]{6.25+\frac{3325}{108}}(\cos(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3})+i\sin(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3}))+(\frac{-1-i\sqrt3}{2})\sqrt[6]{6.25+\frac{3325}{108}}(\cos(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3})-i\sin(\frac{\tan^{-1}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3}))$$

We apply FOIL to multiply complex numbers.

$$a_2=-\frac12\sqrt[6]{6.25+\frac{3325}{108}}(\cos(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3})-\frac{i}{2}\sqrt[6]{6.25+\frac{3325}{108}}\sin(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3}))+\frac{i\sqrt3}{2}\sqrt[6]{6.25+\frac{3325}{108}}(\cos(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3})-\frac{\sqrt3}{2}\sqrt[6]{6.25+\frac{3325}{108}}(\sin(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3}))-\frac12\sqrt[6]{6.25+\frac{3325}{108}}(\cos(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3})+\frac{i}{2}\sqrt[6]{6.25+\frac{3325}{108}}\sin(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3}))-\frac{i\sqrt3}{2}\sqrt[6]{6.25+\frac{3325}{108}}(\cos(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3})-\frac{\sqrt3}{2}\sqrt[6]{6.25+\frac{3325}{108}}(\sin(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3})) $$

Note that the sum of the imaginary pasrts is equal to zero, so $a_2$ is real.

$$a_2=-\sqrt[6]{6.25+\frac{3325}{108}}(\cos(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3})-\sqrt3\sqrt[6]{6.25+\frac{3325}{108}}(\sin(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3})) $$

this is about $-3.3876$

Now we let $w_1=\frac{-1-i\sqrt3}{2}$, and $w_2=\frac{-1+i\sqrt3}{2}$

This should also have a real result.

$$a_3=-\sqrt[6]{6.25+\frac{3325}{108}}(\cos(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3})+\sqrt3\sqrt[6]{6.25+\frac{3325}{108}}(\sin(\frac{\tan^{-1 rad}(\frac{\sqrt3325}{15\sqrt3})+\pi}{3}))$$

This is about $2.8741$

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  • $\begingroup$ It just occureed to me. If a cubic equation over the reals have three real roots, then they come in the form of: $$r_1+2r_2\cos(\theta)$$ $$r_1-r_2\cos(\theta)-(\sqrt3)r_2\sin(\theta)$$ $$r_1-r_2\cos(\theta)+(\sqrt3)r_2\sin(\theta)$$ $\endgroup$ Mar 7 at 18:41

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