Answer:
[tex]\frac{7}{2v^6\left(7v-1\right)}[/tex]
Step-by-step explanation:
So lets start with a step by step process in how we come to this answer.
Using the format of [tex]\frac{v^{-6}}{7v\cdot \:6-6}[/tex] we can then [tex]\mathrm{Multiply\:the\:numbers:}\:7\cdot \:6=42[/tex] which thus gives us the result of [tex]\frac{v^{-6}}{42v-6}[/tex]. Now we want to apply the exponent rule of [tex]a^{-b}=\frac{1}{a^b}[/tex] which for our equation is [tex]v^{-6}=\frac{1}{v^6}[/tex]. With that in mind we can now apply this to our current result of [tex]\frac{v^{-6}}{42v-6} > \frac{\frac{1}{v^6}}{42v-6}[/tex]. Now we want to apply the fraction rule [tex]\frac{\frac{b}{c}}{a}=\frac{b}{c\:\cdot \:a}[/tex]. This gives us [tex]\frac{1}{v^6\left(42v-6\right)}[/tex] which then gives us [tex]21\cdot \frac{1}{v^6\left(42v-6\right)}[/tex]. Use the multiplicity rule of [tex]a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}[/tex] to get [tex]\frac{1\cdot \:21}{v^6\left(42v-6\right)}[/tex]. Now multiply 1 by 2 to get [tex]\frac{21}{v^6\left(42v-6\right)}[/tex]. Moving on from this, we now must factor 42v - 6. [tex]\mathrm{Rewrite\:as} > 6\cdot \:7v-6\cdot \:1 > \mathrm{Factor\:out\:common\:term\:}6 > 6\left(7v-1\right)[/tex]. Now we can apply this result to our previous result and get [tex]\frac{21}{v^6\cdot \:6\left(7v-1\right)}[/tex]. All that is left is to cancel the common factor of 3 to get [tex]\frac{7}{2v^6\left(7v-1\right)}[/tex]. Hope this helps!
