Question

In two or three sentences explain how you would solve for the real solutions of the following equation: please help asap!!

Answer 1

the real solutions for the equation $x^{3}=-64$ are -

$x=4,2+-2\sqrt{3i}$

Step-by-step explanation:

   $x^{3}$ = $- 64$

   $x^{3} +64$  = 0

   We can write 64 as  $4^{3}$

  $x^{3}$ + $4^{3}$ = 0

  using the identity  ( $x^{3}+y^{3} = (x+y)(x^{2} -xy+y^{2} )$ )

we get,

  = $(x+4) (x^{2} -x*4+4^{2} )$

  = $(x+4)(x^{2} -4x+16)$    ....................(1)

 solving the quadratic equation  ,

   $x^{2} -4x+16$ =0

solutions of this quadratic equation can be obtained by

   $x=-b +- \sqrt{b^{2}-4ac } /2a$

let use y for factors

$x=-(-4x)+-\sqrt{(-4x^{2} )-4*x^{2} *16} / 2*x^{2}$

$x=4x+-\sqrt{16x^{2} -64x^{2} } /2x^{2}$

$x=4+-\sqrt{16-64}/2$

$x=4+-4\sqrt{3i} /2$

<u />$x=2+-2\sqrt{3i}$    ..................(2)

from the equation 1 we have,

$x-4=0$

which gives solution $x=4$

and from equation 2 we got  $x=2+-2\sqrt{3i}$

so the real solutions for the equation $x^{3}=-64$ are -

$x=4,2+-2\sqrt{3i}$