What is (3y^3)^4(4y^2)^{-3} ?

The solution to (3y^3)^4(4y^2)^{-3} is (81y^6)/(64)

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simplify (3y^3)^4(4y^2)^{-3}

simplify

(3 y^{3})^{4} (4 y^{2})^{- 3}

\frac{81 y ^{6}}{64}

Solution steps

(3 y^{3})^{4} (4 y^{2})^{- 3}

Simplify

(3 y^{3})^{4} : 81 y^{12}

= 81 y^{12} (4 y^{2})^{- 3}

Simplify

(4 y^{2})^{- 3} : \frac{1}{64 y ^{6}}

= 81 y^{12} \frac{1}{64 y ^{6}}

Apply the fraction rule:

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

= \frac{81 y ^{12} \cdot 1}{64 y ^{6}}

Multiply the numbers:

81 \cdot 1 = 81

= \frac{81 y ^{12}}{64 y ^{6}}

Simplify

\frac{y ^{12}}{y ^{6}} : y^{6}

= \frac{81 y ^{6}}{64}

(a^{4})^{2}

7 < x + 9 < 9

∣ x - 3 ∣ - ∣ x + 1 ∣ > 2

x \geq - 1 and x \leq 8

w^{2} - 10 w + 24 = 0

What is (3y^3)^4(4y^2)^{-3} ?
The solution to (3y^3)^4(4y^2)^{-3} is (81y^6)/(64)