<PRE><B>Question 177739</B>
{{{A=x^3}}}....eq 1
:
{{{(x+12)(x+6)(x-4)=2A}}}...eq 2
:
take value of A from eq 1 and plug it into eq 2
:
{{{(x+12)(x+6)(x-4)=2x^3}}}
:
{{{(x^2+18x+72)(x-4)=2x^3}}}
:
{{{x^3-4x^2+18x^2-72x+72x-288=2x^3}}}
:
{{{x^3-14x^2+288=0}}}

if we take the factors of plus or minus 288/1 we have possible rational roots
:
if you plug 6 in you will find that to be a factor so lets use synthetic division to factor 6 out of our equation
:
:
  l 1  -14   0      288
  l
6 l____  6___-48___-288____________________________
    1   -8    -48     0

so we have (x-6)(x^2-8x-48)
which can be further factored to (x-6)(x-12)(x+4)
:
so x is 6,12,0r -4..since 4 is negative we throw it out
:
{{{system(x=6,x=12)}}}
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