Question 189584
Can you please help me solve this? I'm supposed to solve for all roots but I'm not sure how. {{{8x^3+27=0}}}
<pre><font size = 4 color = "indigo"><b>
You must learn the principle of factoring the sum and difference of
two cubes:

{{{FIRST^3 + SECOND^3}}} 

factors as

{{{(FIRST+SECOND)(FIRST^2-FIRST*SECOND+SECOND^2)}}}

and

{{{FIRST^3 - SECOND^3}}} 

factors as

{{{(FIRST-SECOND)(FIRST^2+FIRST*SECOND+SECOND^2)}}}

------------------------------

Your problem is the first case as it has a + sign between
the two terms:

{{{8x^3+27=0}}}

Notice that {{{8x^3}}} can be written as {{{(2x)^3}}}

Notice that {{{27}}} can be written as {{{(3)^3}}}

So we write it as 

{{{(2x)^3+(3)^3=0}}}

So {{{FIRST=2x}}} and {{{SECOND=3}}}

and since

{{{FIRST^3 + SECOND^3}}} 

factors as

{{{(FIRST+SECOND)(FIRST^2-FIRST*SECOND+SECOND^2)}}}

we factor the left side of 

{{{(2x)^3+(3)^3=0}}}

as

{{{(2x+3)((2x)^2-(2x)*(3)+(3)^2)=0)}}}

Simplifying inside the second parentheses:

{{{(2x+3)((2^2)(x^2)-(6x)+9)=0}}}

{{{(2x+3)(4x^2-6x+9)=0}}}

So we use the zero factor property:

{{{2x+3=0}}},  {{{4x^2-6x+9=0}}}

Solving the first,

{{{2x+3=0}}}
{{{2x=-3}}}
{{{x=-3/2}}}

So one root is {{{-3/2}}} 

Solving the second:

{{{4x^2-6x+9=0}}}

That doesn't factor so we resort to the
quadratic formula:

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

where {{{a=4}}}, {{{b=-6}}}, {{{c=9}}}

{{{x = (-(-6) +- sqrt( (-6)^2-4*(4)*(9) ))/(2*(4)) }}} 

{{{x = (6 +- sqrt(36-144) )/8 }}} 

{{{x = (6 +- sqrt(-108) )/8 }}}

{{{x = (6 +- i*sqrt(108) )/8 }}}

{{{x = (6 +- i*sqrt(36*3) )/8 }}}

{{{x = (6 +- 6i*sqrt(3) )/8 }}}

Make two fractions on the right

{{{x = 6/8 +- (6i*sqrt(3))/8}}}

Reduce the fractions by dividing top and 
bottom by 2:

{{{x = 3/4 +- (3i*sqrt(3))/4}}}

So the other two roots are

{{{3/4 + (3sqrt(3))/4}}}{{{i}}}

and 

{{{3/4 - (3sqrt(3))/4}}}{{{i}}}

Edwin</pre>