Question 371391
{{{(42b^9-2b^7+48b^3)/(6b^3-24b^7)}}}
First we factor out the Greatest Common Factor (GCF). The GCF in the numerator is {{{2b^3}}} and in the denominator it is {{{6b^3}}}:
{{{(2b^3(21b^6-b^4+24))/(6b^3(1-4b^4))}}}
The second factor of the numerator will not factor. So even though the second factor in the denominator is a difference of squares and will factor, there is no point because neither of those factors will cancel with a factor in the numerator. All we can do now is factor the 6:
{{{(2b^3(21b^6-b^4+24))/(2*3*b^3(1-4b^4))}}}
Now we can cancel:
{{{(cross(2)cross(b^3)(21b^6-b^4+24))/(cross(2)*3*cross(b^3)(1-4b^4))}}}
leaving:
{{{(21b^6-b^4+24)/(3*(1-4b^4))}}}
or
{{{(21b^6-b^4+24)/(3-12b^4))}}}