Question

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

Answer 1

The roots are:

`x_n = 1/24(7+2sqrt(1513) cos(1/3 cos^(-1)((10531 sqrt(1513))/2289169) + (2npi)/3))`

for `n = 0, 1, 2`

Explanation:

`8x^3-7x^2-61x+6 = 0`

`color(white)()`
Discriminant

The discriminant `Delta` of a cubic polynomial in the form `ax^3+bx^2+cx+d` is given by the formula:

`Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd`

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=1728f(x)=13824x^3-12096x^2-105408x+10368`

`=(24x-7)^3-4539(24x-7)-21062`

`=t^3-4539t-21062`

where `t=(24x-7)`

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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:

`4cos^3 theta - 3 cos theta = cos 3 theta`

Let `t = k cos theta`

`0 = t^3-4539t-21062`

`color(white)(0) = k^3 cos^3 theta - 4539 k cos theta - 21062`

`color(white)(0) = (k^3/4) 4cos^3 theta - (1513 k) 3cos theta - 21062`

So we want:

`k^3/4 = 1513k`

Hence:

`k^2 = 4*1513`

So:

`k = +-2sqrt(1513)`

Let `k = 2sqrt(1513)`

Then `1513k = 3026sqrt(1513)`

Then:

`0 = (k^3/4) 4cos^3 theta - (1513 k) 3cos theta - 21062`

`color(white)(0) = 3026sqrt(1513)(4cos^3 theta - 3cos theta) - 21062`

`color(white)(0) = 3026sqrt(1513)(cos 3 theta) - 21062`

Hence:

`cos 3 theta = 21062/(3026 sqrt(1513)) = (10531 sqrt(1513))/2289169`

So:

`3 theta = +-cos^(-1)((10531 sqrt(1513))/2289169) + 2npi`

`theta = +- 1/3 cos^(-1)((10531 sqrt(1513))/2289169) + (2npi)/3`

Hence:

`t_n = 2sqrt(1513) cos(1/3 cos^(-1)((10531 sqrt(1513))/2289169) + (2npi)/3)`

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

Then `x = 1/24(t + 7)`

So we have roots:

`x_n = 1/24(7+2sqrt(1513) cos(1/3 cos^(-1)((10531 sqrt(1513))/2289169) + (2npi)/3))`

for `n = 0, 1, 2`