What is ((3x^5y^3)/(x^3y^6))^4 ?

The solution to ((3x^5y^3)/(x^3y^6))^4 is (81x^8)/(y^{12)}

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simplify ((3x^5y^3)/(x^3y^6))^4

simplify

(\frac{3 x ^{5} y ^{3}}{x ^{3} y ^{6}})^{4}

\frac{81 x ^{8}}{y ^{12}}

Solution steps

(\frac{3 x ^{5} y ^{3}}{x ^{3} y ^{6}})^{4}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( 3 x ^{5} y ^{3} ) ^{4}}{( x ^{3} y ^{6} ) ^{4}}

Simplify

(3 x^{5} y^{3})^{4} : 81 x^{20} y^{12}

= \frac{81 x ^{20} y ^{12}}{( x ^{3} y ^{6} ) ^{4}}

Simplify

(x^{3} y^{6})^{4} : x^{12} y^{24}

= \frac{81 x ^{20} y ^{12}}{x ^{12} y ^{24}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

\frac{x ^{20}}{x ^{12}} = x^{20 - 12}

= \frac{81 y ^{12} x ^{20 - 12}}{y ^{24}}

Subtract the numbers:

20 - 12 = 8

= \frac{81 x ^{8} y ^{12}}{y ^{24}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{y ^{12}}{y ^{24}} = \frac{1}{y ^{24 - 12}}

= \frac{81 x ^{8}}{y ^{24 - 12}}

Subtract the numbers:

24 - 12 = 12

= \frac{81 x ^{8}}{y ^{12}}

\frac{9 + ( - 3 )}{2}

- x^{2} + 6 x + 3 = 0

\frac{x + 2}{x ^{2} - 3 x - 10}

4 x^{2} - 33 x + 63

x^{2} - x - 11 = 0

What is ((3x^5y^3)/(x^3y^6))^4 ?
The solution to ((3x^5y^3)/(x^3y^6))^4 is (81x^8)/(y^{12)}