Mohammad Cannon

2022-06-24

simplify fractions with exponent

The given fraction is: $(\frac{a}{b}{)}^{n}\cdot (\frac{b}{c}{)}^{n}\cdot (\frac{c}{a}{)}^{n+1}$

The given solution is: $\frac{c}{a}$

What I have done so far:

$(\frac{a}{b}{)}^{n}\cdot (\frac{b}{c}{)}^{n}\cdot (\frac{c}{a}{)}^{n+1}$ | multiply $\frac{a}{b}$ and $\frac{b}{c}$ because of same exponent

$(\frac{ab}{bc}{)}^{n}\ast (\frac{c}{a}{)}^{n+1}$ | get rid of $b$

$(\frac{a}{c}{)}^{n}\ast (\frac{c}{a}{)}^{n+1}$

Can you please explain how I continue simplifying or what I did wrong? Thanks!

The given fraction is: $(\frac{a}{b}{)}^{n}\cdot (\frac{b}{c}{)}^{n}\cdot (\frac{c}{a}{)}^{n+1}$

The given solution is: $\frac{c}{a}$

What I have done so far:

$(\frac{a}{b}{)}^{n}\cdot (\frac{b}{c}{)}^{n}\cdot (\frac{c}{a}{)}^{n+1}$ | multiply $\frac{a}{b}$ and $\frac{b}{c}$ because of same exponent

$(\frac{ab}{bc}{)}^{n}\ast (\frac{c}{a}{)}^{n+1}$ | get rid of $b$

$(\frac{a}{c}{)}^{n}\ast (\frac{c}{a}{)}^{n+1}$

Can you please explain how I continue simplifying or what I did wrong? Thanks!

Alisa Gilmore

Beginner2022-06-25Added 22 answers

$(\frac{a}{b}{)}^{n}\cdot (\frac{b}{c}{)}^{n}\cdot (\frac{c}{a}{)}^{n+1}=$

$\frac{{a}^{n}}{{b}^{n}}\cdot \frac{{b}^{n}}{{c}^{n}}\cdot \frac{{c}^{n+1}}{{a}^{n+1}}=$

$\frac{{a}^{n}{b}^{n}{c}^{n+1}}{{b}^{n}{c}^{n}{a}^{n+1}}=$

$\frac{{b}^{n}c}{{b}^{n}a}=$

$\frac{c}{a}$

$\frac{{a}^{n}}{{b}^{n}}\cdot \frac{{b}^{n}}{{c}^{n}}\cdot \frac{{c}^{n+1}}{{a}^{n+1}}=$

$\frac{{a}^{n}{b}^{n}{c}^{n+1}}{{b}^{n}{c}^{n}{a}^{n+1}}=$

$\frac{{b}^{n}c}{{b}^{n}a}=$

$\frac{c}{a}$

boloman0z

Beginner2022-06-26Added 10 answers

You're almost there:

Note that:

${\left(\frac{c}{a}\right)}^{n+1}={\left(\frac{c}{a}\right)}^{n}{\left(\frac{c}{a}\right)}^{1}$

Your expression becomes:

${\left(\frac{abc}{abc}\right)}^{n}\frac{c}{a}$

Does that help?

Note that:

${\left(\frac{c}{a}\right)}^{n+1}={\left(\frac{c}{a}\right)}^{n}{\left(\frac{c}{a}\right)}^{1}$

Your expression becomes:

${\left(\frac{abc}{abc}\right)}^{n}\frac{c}{a}$

Does that help?

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