# Powers and Roots – Simplify an expression with powers – Exercise 1660

## Exercise

Simplify the expression:

$$\frac{{(a^2 a^3)}^n\cdot a^{n+7}\cdot {(b^4)}^{n+3}}{{{({(a^2)}^{n+1}\cdot b^{n+1}})}^3\cdot b^{n+8}}$$

$$\frac{{(a^2 a^3)}^n\cdot a^{n+7}\cdot {(b^4)}^{n+3}}{{{({(a^2)}^{n+1}\cdot b^{n+1}})}^3\cdot b^{n+8}}=ab$$

## Solution

Using Powers and Roots rules we get:

$$\frac{{(a^2 a^3)}^n\cdot a^{n+7}\cdot {(b^4)}^{n+3}}{{{({(a^2)}^{n+1}\cdot b^{n+1}})}^3\cdot b^{n+8}}=$$

$$=\frac{{(a^{2+3})}^n\cdot a^{n+7}\cdot b^{4(n+3)}}{{{(a^{2(n+1)}\cdot b^{n+1}})}^3\cdot b^{n+8}}=$$

$$=\frac{{(a^5)}^n\cdot a^{n+7}\cdot b^{4n+12}}{{a^{3(2n+2)}\cdot b^{3(n+1)}}\cdot b^{n+8}}=$$

$$=\frac{a^{5n}\cdot a^{n+7}\cdot b^{4n+12}}{a^{6n+6}\cdot b^{3(n+1)+(n+8)}}=$$

$$=\frac{a^{5n+(n+7)}\cdot b^{4n+12}}{a^{6n+6}\cdot b^{3n+3+n+8}}=$$

$$=\frac{a^{6n+7}\cdot b^{4n+12}}{a^{6n+6}\cdot b^{4n+11}}=$$

$$=a^{(6n+7)-(6n+6)}\cdot b^{(4n+12)-(4n+11)}=$$

$$=a^{(6n+7-6n-6}\cdot b^{4n+12-4n-11}=$$

$$=ab$$

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