Simplification of an algebraic expression

[Guest]

How could I simplify this?

(-3a^n*b^(2n-2)^3) * 4 (ab^1-n)^2
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(2*a^n+2*b^3)^6

 

[tkhunny]

One piece at a time.

First, what is that ab^1-n?? Add some parentheses to clarify that, then show us your first step.

 

[soroban]

Hello, americo74!

I'll take a guess at what you meant . . .

Simplify: \(\displaystyle \L\:\frac{\left(-3\cdot a^n\cdot b^{2n-2}\right)^3\,\cdot\,4\left(a\cdot b^{1-n})^2} {\left(2a^{n+2}\cdot b^3\right)^6}\)

We have: \(\displaystyle \L\:\frac{(-3)^3\cdot\left(a^n\right)^3\cdot\left(b^{2n-2}\right)^3\cdot4\cdot a^2\cdot\left(b^{1-n}\right)^2} {2^6\cdot\left(a^{n+2}\right)^6\cdot\left(b^3\right)^6}\)

. . \(\displaystyle \L=\;\frac{-27\,\cdot\,a^{3n}\,\cdot\,b^{6n-6}\,\cdot\,4\,\cdot\,a^2\,\cdot\,b^{2-2n}} {64\,\cdot\,a^{6n+12}\,\cdot\,b^{18}}\)

. . \(\displaystyle \L=\;\frac{-27\cdot4}{64}\,\cdot\,\frac{a^{3n}\,\cdot\,a^2}{a^{6n+12}} \,\cdot\,\frac{b^{6n-6}\,\cdot\,b^{2-2n}}{b^{18}}\)

. . \(\displaystyle \L=\;-\frac{27}{16}\,\cdot\,\frac{a^{3n+2}}{a^{6n+12}}\,\cdot\,\frac{b^{4n-4}}{b^{18}}\)

. . \(\displaystyle \L=\;-\frac{27\,b^{^{4n-22}}}{16\,a^{^{3n+10}}}\)