Question

solve x^3-26x+60=0 by cardan's method?​

Answer 1

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x³-26x+60=0

x³-26x=-60

x(x²-26)=-60

x=-60,

x²-26=-60

x²-26+60=0

x²+34=0

x=√-34

x=√34i

Answer 2

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

$f(x) = {x}^{3} - 26x + 60 = 0$

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

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

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

NOW,

$\sf : = x = u + v$

$\sf \implies {(u + v)}^{3} - 26(u + v) + 60 = 0$

$\sf ( {u}^{3} + {v}^{3} + 60) + (u + v)(3uv - 26) = 0$

Therefore, Simultaneously:

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

$\sf \: uv = \frac{26}{3}$

Multiplying First equation by 'u³'

$\sf {u}^{6} + 60 {u}^{3} + (\frac{26}{3} )^{3} = 0$

$\sf {u}^{3} = \frac{ - 270 \pm82 \sqrt{3} }{9}$

But, as u and v are indistinguishable

$\sf\therefore {u}^{3} = \frac{ - 270 + 82 \sqrt{3} }{9} \\ \: \: \: \: \sf {v}^{3} = \frac{ - 270 - 82 \sqrt{3} }{9}$

$\sf u = \omega _{k} \sqrt[3]{ \frac{ - 270 + 82 \sqrt{3} }{9} }$

$\sf v = \omega _{k} ^{2} \sqrt[3]{ \frac{ - 270 - 82 \sqrt{3} }{9} }$

OR

$\sf u = \omega _{k}\frac{ \sqrt[3]{ - 810 + 246 \sqrt{3} } }{3} \\ \sf v = \omega _{k}^{2} \frac{ \sqrt[3]{ - 810 - 246 \sqrt{3} } }{3}$

After Simplification

(The Number inside the cube roots comes out to be a perfect cube)

$\sf u = \omega_k \frac{ \sqrt{3} - 9}{3} \\ \: \: \: \: \sf v = - \omega_k^2 \frac{ \sqrt{3} + 9}{3}$

Therefore,

The three solutions of the cubic equation are:

$\bf x_1 = - 6$

$\bf{ x_{2} = 3 + } \it{i}$

$\bf{x _{3} = 3 - } \it{i}$

where, i = √(–1)