Question

Simplify 21v^-6/(7v6-6)

Answer 1

`(7)/(2v^6\left(7v-1\right))`

Explanation:

So lets start with a step by step process in how we come to this answer.

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

Answer 2

-3 on edge:)

Explanation: