Question 59095
Formula for the factoring of the difference of cubes:{{{highlight(u^3-v^3=(u-v)(u^2+uv+v^2))}}}
{{{27x^3-8=0}}}  
{{{(3x)^3-(2)^3=0}}}  u=3x and v=2
{{{(3x-2)((3x)^2+(3x)(2)+(2)^2)=0}}}
{{{(3x-2)(9x^2+6x+4)=0}}}
{{{3x-2=0}}} and {{{9x^2+6x+4=0}}}
{{{3x-2+2=0+2}}}
{{{3x=2}}}
{{{3x/3=2/3}}}
{{{highlight(x=2/3)}}}  This is the real solution that we found before.  You have to use the quadratic formula to find the complex solution.
  When a quadratic equation is in standard form:{{{highlight(0=ax^2+bx+c)}}} the quadratic formula can be applied: {{{highlight(x=(-b+-sqrt(b^2-4ac))/(2a))}}}
For {{{9x^2+6x+4=0}}}  a=9, b=6 and c=4
{{{x=(-(6)+-sqrt((6)^2-4(9)(4)))/(2(9))}}}
{{{x=(-6+-sqrt(36-144))/18}}}
{{{x=(-6+-sqrt(-108))/18}}}
{{{x=(-6+-sqrt(-1)*sqrt(36)*sqrt(3))/18}}}
{{{x=(-6+or-6*sqrt(3)*i)/18}}}
{{{x=6(-1+-sqrt(3)*i)/(6*3)}}}
{{{x=(-1+-sqrt(3)*i)/3}}}
Happy Calculating!!!


Happy Calculating!!!