Question 1165761
a^3+b^3 factors into (a+b) (a^2-ab+b^2)

so {(x-3)+(3x-2)}* {(x-3)^2)-(x-3)(3x-2)+(3x-2)^2}

the other is
{(x3/8)-8} * ((x^6/64)+x^3+64)}
but the first can be factored as well
{(x/2)-2} * {(x^2/4)+x+4}

and the whole thing is
{(x/2)-2} * {(x^2/4)+x+4}*{((x^6/64)+x^3+64)}