Question

solve x^3-27x+60=0 by cardan's method?​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

$\bf \red{\underline{To \: Find :}} \blue{values \: of \: x \: such \: that}$

$f(x) = {x}^{3} - 27x + 60 = 0$

$\star \star \green{ \bf \underline{Solution}} \star \star$

$\bf{As}, \sf{ Degree[f(x),x] = 3} \\ and, \\ \sf Coefficient [ f(x), {x}^{2} ] = 0$

$\sf \implies x \: is \: a \: binomial$

NOW,

$\sf : = x = u + v$

$\sf \implies {(u + v)}^{3} - 27(u + v) + 60 = 0$

$\sf ( {u}^{3} + {v}^{3} + 60) + (u + v)(3uv - 27) = 0$

Therefore, Simultaneously:

$\sf {v}^{3} + 60 + {u}^{3} = 0$

$\sf \: uv = 9$

Multiplying First equation by 'v³'

$\sf {v}^{6} + 60 {v}^{3} + (9)^{3} = 0$

$\sf {v}^{3} = - 30 \pm3 \sqrt{19}$

But, at the same time u and v are indistinguishable

$\sf u = \omega _{k} \sqrt[3]{ - 30 + 3 \sqrt{19} } \\ \sf \: v = \omega _{k} {}^{2} \sqrt[3]{ - 30 - 3 \sqrt{19} }$

Therefore, the three solutions are:

$\bf x_1= \sqrt[3]{ - 30 + 3 \sqrt{19} } + \sqrt[3]{ - 30 - \sqrt{19} }$

$\bf x_{2,3} = \frac{ \sqrt[3]{30 + 3\sqrt{19} } + \sqrt[3]{30 - 3 \sqrt{19} } }{2} \pm \it{i} \bf \frac{ \sqrt{3} }{2}( \sqrt[3]{30 + 3 \sqrt{19} } - \sqrt[3]{30 - 3 \sqrt{19} } \: )$