Question

x^3-0x^2-53x-42=?how solve this

Answer 1

x³ - 53 x - 42 = 0
solutions can be obtained using  standard formulae
       x = -6.84579

Cubic equation / cubic polynomial  :  x³ - 3 p x - 2 q = 0
Solution is given by:    
$x_k,\ \ k=0,1,2\\\\x_k=2\ *\ \sqrt{p}\ *\ Cos[\frac{1}{3}\ *\ Cos^{-1}(\frac{q}{p^{\frac{3}{2}}})-\frac{2\pi k}{3}]$

Or, by:
$x = [q+\sqrt{q^2-p^3}]^{\frac{1}{3}}+[q-\sqrt{q^2-p^3}]^{\frac{1}{3}}$

By substituting  p = 53/3      and q = 21  we get:
$x = [ 21 + 71.2247 i ]^{\frac{1}{3}} + [ 21 - 71.2247 i ]^{\frac{1}{3}}$
This formula yields in complex numbers. A little difficult to solve.  we have to represent complex numbers in polar form and then take cube roots and add the two cube roots.

P = 53/3 and q = 21: so

$x_0=2*\sqrt{53/3}*Cos[\frac{1}{3}*Cos^{-1}(\frac{21}{(53/3)^{\frac{3}{2}}})-\frac{2\pi*0}{3}]\\=7.647983\\\\x_1=2*\sqrt{53/3}*Cos[\frac{1}{3}*Cos^{-1}(\frac{21}{(53/3)^{\frac{3}{2}}})-\frac{2\pi*1}{3}]\\=-0.8021928\\\\x_2=2*\sqrt{53/3}*Cos[\frac{1}{3}*Cos^{-1}(\frac{21}{(53/3)^{\frac{3}{2}}})-\frac{2\pi*2}{3}]\\=-6.84579$

These are the three roots of the given polynomial.