The expression $729x^3+8$ can be written as $(ax+b)(cx^2+dx+e)$. Find $a+b+c+d+e$.
To factor $729x^3+8$, I notice that it is the sum of two cubes: $729x^3=(9x)^3$ and $8=2^3$. There is a formula for factoring the sum of two cubes: $a^3+b^3=(a+b)(a^2-ab+b^2)$. Using this formula, I can write $729x^3+8=(9x+2)((9x)^2-9x\cdot 2+2^2)$. Expanding the second factor, I get $(9x+2)(81x^2-18x+4)$. Now I can identify the coefficients $a=9$, $b=2$, $c=81$, $d=-18$, and $e=4$. To find their sum, I just add them up: $a+b+c+d+e=9+2+81-18+4=78$. 