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

The solution to (4y^{-3})^2 is (16)/(y^6)

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

simplify

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

\frac{16}{y ^{6}}

Solution steps

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

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

4^{2} = 16

= 16 (y^{- 3})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c},

assuming

a \geq 0

(y^{- 3})^{2} = y^{- 3 \cdot 2}

= 16 y^{- 3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= 16 y^{- 6}

Apply exponent rule:

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

y^{- 6} = \frac{1}{y ^{6}}

= 16 \cdot \frac{1}{y ^{6}}

Simplify

16 \cdot \frac{1}{y ^{6}} : \frac{16}{y ^{6}}

= \frac{16}{y ^{6}}

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