# What are the roots of 8x^3-7x^2-61x+6=0 ?

Sep 25, 2016

The roots are:

${x}_{n} = \frac{1}{24} \left(7 + 2 \sqrt{1513} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{10531 \sqrt{1513}}{2289169}\right) + \frac{2 n \pi}{3}\right)\right)$

for $n = 0 , 1 , 2$

#### Explanation:

$8 {x}^{3} - 7 {x}^{2} - 61 x + 6 = 0$

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Discriminant

The discriminant $\Delta$ of a cubic polynomial in the form $a {x}^{3} + b {x}^{2} + c x + d$ is given by the formula:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

In our example, $a = 8$, $b = - 7$, $c = - 61$ and $d = 6$, so we find:

$\Delta = 182329 + 7263392 + 8232 - 62208 + 368928 = 7760673$

Since $\Delta > 0$ this cubic has $3$ Real zeros.

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Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

$0 = 1728 f \left(x\right) = 13824 {x}^{3} - 12096 {x}^{2} - 105408 x + 10368$

$= {\left(24 x - 7\right)}^{3} - 4539 \left(24 x - 7\right) - 21062$

$= {t}^{3} - 4539 t - 21062$

where $t = \left(24 x - 7\right)$

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Trigonometric substitution

Look for a substitution of the form $t = k \cos \theta$ where $k$ is chosen to cause the resulting equation to contain:

$4 {\cos}^{3} \theta - 3 \cos \theta = \cos 3 \theta$

Let $t = k \cos \theta$

$0 = {t}^{3} - 4539 t - 21062$

$\textcolor{w h i t e}{0} = {k}^{3} {\cos}^{3} \theta - 4539 k \cos \theta - 21062$

$\textcolor{w h i t e}{0} = \left({k}^{3} / 4\right) 4 {\cos}^{3} \theta - \left(1513 k\right) 3 \cos \theta - 21062$

So we want:

${k}^{3} / 4 = 1513 k$

Hence:

${k}^{2} = 4 \cdot 1513$

So:

$k = \pm 2 \sqrt{1513}$

Let $k = 2 \sqrt{1513}$

Then $1513 k = 3026 \sqrt{1513}$

Then:

$0 = \left({k}^{3} / 4\right) 4 {\cos}^{3} \theta - \left(1513 k\right) 3 \cos \theta - 21062$

$\textcolor{w h i t e}{0} = 3026 \sqrt{1513} \left(4 {\cos}^{3} \theta - 3 \cos \theta\right) - 21062$

$\textcolor{w h i t e}{0} = 3026 \sqrt{1513} \left(\cos 3 \theta\right) - 21062$

Hence:

$\cos 3 \theta = \frac{21062}{3026 \sqrt{1513}} = \frac{10531 \sqrt{1513}}{2289169}$

So:

$3 \theta = \pm {\cos}^{- 1} \left(\frac{10531 \sqrt{1513}}{2289169}\right) + 2 n \pi$

$\theta = \pm \frac{1}{3} {\cos}^{- 1} \left(\frac{10531 \sqrt{1513}}{2289169}\right) + \frac{2 n \pi}{3}$

Hence:

${t}_{n} = 2 \sqrt{1513} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{10531 \sqrt{1513}}{2289169}\right) + \frac{2 n \pi}{3}\right)$

with distinct values for $n = 0 , 1 , 2$

Then $x = \frac{1}{24} \left(t + 7\right)$

So we have roots:

${x}_{n} = \frac{1}{24} \left(7 + 2 \sqrt{1513} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{10531 \sqrt{1513}}{2289169}\right) + \frac{2 n \pi}{3}\right)\right)$

for $n = 0 , 1 , 2$