What is ((2x^3y^2z)/(3x^4yz^{-2)})^2 ?

The solution to ((2x^3y^2z)/(3x^4yz^{-2)})^2 is (4y^2z^6)/(9x^2)

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simplify ((2x^3y^2z)/(3x^4yz^{-2)})^2

simplify

(\frac{2 x ^{3} y ^{2} z}{3 x ^{4} y z ^{- 2}})^{2}

\frac{4 y ^{2} z ^{6}}{9 x ^{2}}

Solution steps

(\frac{2 x ^{3} y ^{2} z}{3 x ^{4} y z ^{- 2}})^{2}

Apply exponent rule:

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

= \frac{( 2 x ^{3} y ^{2} z ) ^{2}}{( 3 x ^{4} y z ^{- 2} ) ^{2}}

Simplify

(2 x^{3} y^{2} z)^{2} : 4 x^{6} y^{4} z^{2}

= \frac{4 x ^{6} y ^{4} z ^{2}}{( 3 x ^{4} y z ^{- 2} ) ^{2}}

Simplify

(3 x^{4} y z^{- 2})^{2} : \frac{9 x ^{8} y ^{2}}{z ^{4}}

= \frac{4 x ^{6} y ^{4} z ^{2}}{\frac{9 x ^{8} y ^{2}}{z ^{4}}}

Apply the fraction rule:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

= \frac{4 x ^{6} y ^{4} z ^{2} z ^{4}}{9 x ^{8} y ^{2}}

Apply exponent rule:

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

\frac{x ^{6}}{x ^{8}} = \frac{1}{x ^{8 - 6}}

= \frac{4 y ^{4} z ^{4} z ^{2}}{9 y ^{2} x ^{8 - 6}}

Subtract the numbers:

8 - 6 = 2

= \frac{4 y ^{4} z ^{4} z ^{2}}{9 x ^{2} y ^{2}}

Apply exponent rule:

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

\frac{y ^{4}}{y ^{2}} = y^{4 - 2}

= \frac{4 z ^{4} z ^{2} y ^{4 - 2}}{9 x ^{2}}

Subtract the numbers:

4 - 2 = 2

= \frac{4 y ^{2} z ^{4} z ^{2}}{9 x ^{2}}

4 y^{2} z^{4} z^{2} = 4 y^{2} z^{6}

= \frac{4 y ^{2} z ^{6}}{9 x ^{2}}

x^{2} + 7 x + \frac{49}{4}

900

2 x (x - 2) + 6 = 2 (x^{2} + 2)

x - 11 = - 5 x^{2}

49 x^{2} + 16 = 56 x

What is ((2x^3y^2z)/(3x^4yz^{-2)})^2 ?
The solution to ((2x^3y^2z)/(3x^4yz^{-2)})^2 is (4y^2z^6)/(9x^2)