Question 611063
{{{(3x^4)^3/x^15}}}
According to the order of operations (aka PEMDAS), exponents come before multiplying or dividing. So we start in the numerator. Using the rule for exponents, {{{(a*b)^q = a^q*b^q}}}, on {{{(3x^4)^3}}} we get:
{{{(3^3*(x^4)^3)/x^15}}}
Using the rule for exponents, {{{(a^q)^p = a^(q*p)}}}, on {{{(x^4)^3}}} and since {{{3^3 = 27}}} we get:
{{{(27x^12)/x^15}}}
Next we use another rule for exponents, {{{a^q/a^p = a^(q-p)}}}, on {{{x^12/x^15}}} we get:
{{{27x^(-3)}}}
This may be an acceptable answer. If you don't want a negative exponent, then we can replace {{{x^(-3)}}} with {{{1/x^3}}}
{{{27(1/x^3)}}}
which simplifies to
{{{27/x^3}}}