the real solutions for the equation [tex]x^{3}=-64[/tex] are -
[tex]x=4,2+-2\sqrt{3i}[/tex]
Step-by-step explanation:
[tex]x^{3}[/tex] = [tex]- 64[/tex]
[tex]x^{3} +64[/tex] = 0
We can write 64 as [tex]4^{3}[/tex]
[tex]x^{3}[/tex] + [tex]4^{3}[/tex] = 0
using the identity ( [tex]x^{3}+y^{3} = (x+y)(x^{2} -xy+y^{2} )[/tex] )
we get,
= [tex](x+4) (x^{2} -x*4+4^{2} )[/tex]
= [tex](x+4)(x^{2} -4x+16)[/tex] ....................(1)
solving the quadratic equation ,
[tex]x^{2} -4x+16[/tex] =0
solutions of this quadratic equation can be obtained by
[tex]x=-b +- \sqrt{b^{2}-4ac } /2a[/tex]
let use y for factors
[tex]x=-(-4x)+-\sqrt{(-4x^{2} )-4*x^{2} *16} / 2*x^{2}[/tex]
[tex]x=4x+-\sqrt{16x^{2} -64x^{2} } /2x^{2}[/tex]
[tex]x=4+-\sqrt{16-64}/2[/tex]
[tex]x=4+-4\sqrt{3i} /2[/tex]
[tex]x=2+-2\sqrt{3i}[/tex] ..................(2)
from the equation 1 we have,
[tex]x-4=0[/tex]
which gives solution [tex]x=4[/tex]
and from equation 2 we got [tex]x=2+-2\sqrt{3i}[/tex]
so the real solutions for the equation [tex]x^{3}=-64[/tex] are -
[tex]x=4,2+-2\sqrt{3i}[/tex]
