First separate numbers and letters in the fraction inside the brackets.

${(\frac{16 w^{8}}{y^{20}})}^{- \frac{3}{4}} = {(16 \times w^{8} \times \frac{1}{y^{20}})}^{- \frac{3}{4}}$

Simplify the parts where possible - in this case only $\frac{1}{y^{20}}$ can be simplified, using the law of indices " $a^{- m} = \frac{1}{a^{m}}$ ".

$\begin{array}{rcl} {(16 \times w^{8} \times \frac{1}{y^{20}})}^{- \frac{3}{4}} & = & {(16 \times w^{8} \times y^{- 20})}^{- \frac{3}{4}} \end{array}$

Use the laws of indices " ${(a b)}^{n} = a^{n} \times a^{m}$ " and " ${(a^{m})}^{n} = a^{m n}$ " to deal with the power of at the end.

$\begin{array}{rcl} {(16 \times w^{8} \times y^{- 20})}^{- \frac{3}{4}} & = & {(16)}^{- \frac{3}{4}} \times {(w^{8})}^{- \frac{3}{4}} \times {(y^{- 20})}^{- \frac{3}{4}} \\ = & {(16)}^{- \frac{3}{4}} \times w^{- 6} \times y^{15} \end{array}$

For $w^{- 6}$ [1]
For $y^{15}$ [1]

Finally evaluate ${(16)}^{- \frac{3}{4}}$ using the law of indices " $a^{\frac{m}{n}} = \sqrt[n]{a^{m}}$ " and " $a^{- m} = \frac{1}{a^{m}}$ ". Rewrite $w^{- 6}$ .

$\begin{array}{rcl} {(16)}^{- \frac{3}{4}} \times w^{- 6} \times y^{15} & = & {(\sqrt[4]{16})}^{- 3} \times \frac{1}{w^{6}} \times y^{15} \\ = & 2^{- 3} \times \frac{1}{w^{6}} \times y^{15} \\ = & \frac{1}{8} \times \frac{1}{w^{6}} \times y^{15} \\ = & \frac{y^{15}}{8 w^{6}} \end{array}$

${(\frac{16 w^{8}}{y^{20}})}^{- \frac{3}{4}} = \frac{y^{15}}{8 w^{6}}$ [1]

$2^{n + 2} + 2^{n + 1}$	$2^{2 n} + 2^{2 (n - 1)}$
$8^{n} + 8^{n - 1}$	$2^{n + 2} + 2^{n - 1}$

	Download	View
Easy
Medium
Hard

Edexcel GCSE Maths

Topic Questions

Algebraic Roots & Indices