Question

x^3-12x+8 by cardan's method
​

Answer 1

Answer:

x^3-3x+12=0

3x^2-3=0

x^2=3/3

x^2=1

x=1

Answer 2

To Find:

values of x such that:

$f(x) = {x}^{3} - 12x + 18 = 0$

As, Degree[f(x) , x] = 3

and, coefficient(x²) = 0

Therefore, x must be a binomial

Therefore, Substituting:

$x = u + v \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : ∀ \: u,v \in ℝ$

And so

${(u + v)}^{3} - 12(u + v) + 18 = 0$

$\\ {u}^{3} + {v}^{3} + 3 uv(u + v) - 12(u + v) + 18 = 0$

$\\ ({u}^{3} + {v}^{3} + 18 )+ (u + v)(3uv - 12) = 0$

Thus, we may conclude that:

(Only Simultaneously)

$\implies {u}^{3} + {v}^{3 } + 18 = 0 \\ \: \: \: \: \: \implies 3uv = 12 ⇔uv = 4$

Mutlitplying the first equation by u³ on both sides

${u}^{6} +18 {u}^{3} + {(uv)}^{3} = 0$

$\implies {u}^{6} + 18 {u}^{3} + 64 = 0$

$\therefore {u}^{3} = \frac{ - 18 \pm \sqrt{324 - 256} }{2}$

$\implies {u}^{3} = - 9 \pm \sqrt{17}$

But, as u and v are indistinguishable

$\therefore u = \sqrt[3]{1} ^{ \neg} \sqrt[3]{ - 9 + \sqrt{17} }$

And,

$v = (\sqrt[3]{1} ^{ \neg} )^{2} \sqrt[3]{ - 9 - \sqrt{17} }$

Thus, all the Roots of the given cubic are:

$\\ x_1 = \sqrt[3]{ - 9 + \sqrt{17} } + \sqrt[3]{ - 9 - \sqrt{17} }$

$\\ x_{2,3} = \frac{ \sqrt[3]{ \: 9 + \sqrt{17} } + \sqrt[3]{ \: 9 - \sqrt{17} } \pm \sqrt{3}i( \sqrt[3]{9 + \sqrt{17} } \: \pm \sqrt[3]{9 - \sqrt{17} } )}{2}$

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