Question 103968
{{{(3x^3)/(x^3+27) - x/(x+3) + (2x^2)/(x^2-3x+9)}}} Start with the given expression




{{{(3x^3)/((x+3)(x^2-3x+9)) - x/(x+3) + (2x^2)/(x^2-3x+9)}}} Factor {{{x^3+27}}} into {{{(x+3)(x^2-3x+9)}}}




So this means the LCM is {{{(x+3)(x^2-3x+9)}}} (since every denominator has either {{{x+3}}}, {{{x^2-3x+9}}}, or both). So we want every fraction to have a denominator of {{{(x+3)(x^2-3x+9)}}}




{{{(3x^3)/((x+3)(x^2-3x+9)) - ((x^2-3x+9)/(x^2-3x+9))(x/(x+3)) + (2x^2)/(x^2-3x+9)}}} Multiply the second fraction by {{{(x^2-3x+9)/(x^2-3x+9)}}}



{{{(3x^3)/((x+3)(x^2-3x+9)) - x(x^2-3x+9)/((x+3)(x^2-3x+9))+ (2x^2)/(x^2-3x+9)}}} Multiply 



{{{(3x^3)/((x+3)(x^2-3x+9)) - (x^3-3x^2+9x)/((x+3)(x^2-3x+9))+ (2x^2)/(x^2-3x+9)}}} Distribute




{{{(3x^3)/((x+3)(x^2-3x+9)) - (x^3-3x^2+9x)/((x+3)(x^2-3x+9))+ ((x+3)/(x+3))((2x^2)/(x^2-3x+9))}}} Multiply the third fraction by {{{(x+3)/(x+3)}}}



{{{(3x^3)/((x+3)(x^2-3x+9)) - (x^3-3x^2+9x)/((x+3)(x^2-3x+9))+ (2x^2)(x+3)/((x+3)(x^2-3x+9))}}} Multiply




{{{(3x^3)/((x+3)(x^2-3x+9)) - (x^3-3x^2+9x)/((x+3)(x^2-3x+9))+ 
(2x^3+6x^2)/((x+3)(x^2-3x+9))}}} Distribute


Since we have a common denominator, we can combine the fractions


{{{(3x^3- (x^3-3x^2+9x)+2x^3+6x^2)/((x+3)(x^2-3x+9)) }}} Combine the fractions



{{{(3x^3-x^3+3x^2-9x+2x^3+6x^2)/((x+3)(x^2-3x+9)) }}} Distribute the negative



{{{(4x^3+9x^2-9x)/((x+3)(x^2-3x+9)) }}} Combine like terms




{{{((x+3)(4x^2-3x))/((x+3)(x^2-3x+9)) }}} Factor the numerator



{{{(cross((x+3))(4x^2-3x))/(cross((x+3))(x^2-3x+9)) }}} Cancel like terms




{{{(4x^2-3x)/(x^2-3x+9) }}} Simplify



{{{(x(4x-3))/(x^2-3x+9) }}} Factor out an x




So {{{(3x^3)/(x^3+27) - x/(x+3) + (2x^2)/(x^2-3x+9)}}} simplifies to {{{(x(4x-3))/(x^2-3x+9) }}}