Question

How to solve a cubic equation algebraically?

Answer 1

formulae ax³+bx2+cx+d=0

Answer 2

$\sf f(x) = a {x}^{3} + b {x}^{2} + cx + d = 0$

The algebraic solution for this cubic must be a trinomial.

The solutions are:

$x_{1,2,3} = - \frac{b}{3a} + \omega_{k} \sqrt[3]{ \frac{9abc - 2 {b}^{3} - 27 {a}^{2}d}{54 {a}^{3} } + \sqrt{ {(\frac{9abc - 2 {b}^{3} - 27 {a}^{2}d}{54 {a}^{3}} )}^{2} + {( \frac{ 3ac - {b}^{2} }{9 {a}^{2} } } ) ^{3} } } + \omega_{k} {}^{2} \sqrt[3]{ \frac{9abc - 2 {b}^{3} - 27 {a}^{2}d}{54 {a}^{3} } - \sqrt{ {(\frac{9abc - 2 {b}^{3} - 27 {a}^{2}d}{54 {a}^{3}} )}^{2} + {( \frac{ 3ac - {b}^{2} }{9 {a}^{2} } } ) ^{3} } }$

One of the simplest Method of solving this cubic is:

So,

$\sf \frac{d ^{2} }{ {dx}^{2} } f(x) = 6ax + 2b = 0$

$\sf \implies x = - \frac{ b}{3a}$

Hence, performing a shift of the inflection point

$\sf {x}^{3} + \frac{b}{a} {x}^{2} + \frac{c}{a} x + \frac{d}{a} = 0$

$\sf(({x + \frac{b}{3a} ) - \frac{b}{3a} ) }^{3} + \frac{b}{a} ((x + \frac{b}{3a} ) - \frac{b}{3a} )^{2} + \frac{c}{a} ((x + \frac{ b}{3a} ) - \frac{b}{3a} ) + \frac{d}{a} = 0$

$\sf(x + \frac{b}{3a} ) {}^{3} + ( \frac{3ac - {b}^{2} }{9 {a}^{2} } )(x + \frac{b}{3a} ) + ( \frac{2 {b}^{3} - 9abc + 27 {a}^{2} d}{27 {a}^{3} } ) = 0$

Letting,

$y = x + \frac{b}{3a} \\ \\ p = \frac{3ac - {b}^{2} }{9 {a}^{2} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: q = \frac{2 {b}^{3} - 9abc + 27 {a}^{2} d}{27 {a}^{3} }$

This reduces us to solve a much simpler cubic equation

$\sf f(y) = {y}^{3} + py + q = 0$

Now, the solution for y due to transformation reduces to a binomial.

Let, the solution be :

$\sf y = u + v$

$\sf \therefore {(u + v)}^{ 3} + p(u + v) + q = 0$

$\sf( {u}^{3} + {v}^{3} + q) + (u + v)(3uv + p) =$

Therefore, Only Simultaneously:

$uv = - \frac{p}{3} ⇔u = - \frac{p}{3}$

${u}^{3} + {v}^{3} + q = 0$

$\implies {v}^{3} + ( \frac{ - p}{3v} )^{3} + q = 0$

${v}^{6} + q {v}^{3} - {( \frac{p}{3}) }^{3} = 0$

Solving using the quadratic formula:

${v}^{3} = - \frac{q}{2} \pm \sqrt{( \frac{ q}{2} ) ^{2} + ( \frac{p}{3} ) ^{3} }$

But, as u and v are indistinguishable, Therefore:

${u}^{3} = - \frac{q}{2} + \sqrt{( \frac{ q}{2} ) ^{2} + ( \frac{p}{3} ) ^{3} }$

${v}^{3} = - \frac{q}{2} - \sqrt{( \frac{ q}{2} ) ^{2} + ( \frac{p}{3} ) ^{3} }$

Hence,

$u = \omega _{k} \sqrt[3]{ - \frac{q}{2} + \sqrt{( \frac{q}{2}) {}^{2} + {( \frac{p}{3} )}^{3} } }$

$v = \omega _{k} {}^{2} \sqrt[3]{ - \frac{q}{2} - \sqrt{( \frac{q}{2}) {}^{2} + {( \frac{p}{3} )}^{3} } }$

Thus, the Three solutions of f(y) are:

$\sf y_1 =\sqrt[3]{ - \frac{q}{2} + \sqrt{( \frac{q}{2}) {}^{2} + {( \frac{p}{3} )}^{3} } } +\sqrt[3]{ - \frac{q}{2} - \sqrt{( \frac{q}{2}) {}^{2} + {( \frac{p}{3} )}^{3} } }$

$\sf y_{2} = \sqrt[3]{ - \frac{q}{16} + \sqrt{( \frac{q}{16}) {}^{2} + {( \frac{p}{12} )}^{3} } } + \sqrt[3]{ - \frac{q}{16} - \sqrt{( \frac{q}{16}) {}^{2} + {( \frac{p}{12} )}^{3} } } + \sqrt{3} i( \sqrt[3]{ \frac{q}{16} + \sqrt{( \frac{q}{16}) {}^{2} + {( \frac{p}{12} )}^{3} } } - \sqrt[3]{ \frac{q}{16} - \sqrt{( \frac{q}{16}) {}^{2} + {( \frac{p}{12} )}^{3} } })$

$\sf y_{3} = \sqrt[3]{ - \frac{q}{16} + \sqrt{( \frac{q}{16}) {}^{2} + {( \frac{p}{12} )}^{3} } } + \sqrt[3]{ - \frac{q}{16} - \sqrt{( \frac{q}{16}) {}^{2} + {( \frac{p}{12} )}^{3} } } - \sqrt{3} i( \sqrt[3]{ \frac{q}{16} + \sqrt{( \frac{q}{16}) {}^{2} + {( \frac{p}{12} )}^{3} } } - \sqrt[3]{ \frac{q}{16} - \sqrt{( \frac{q}{16}) {}^{2} + {( \frac{p}{12} )}^{3} } })$

And Now, we can solve for x

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thank you

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