How do you simplify $(8 b^{2}) (2 b^{- 8}) (2 b^{- 6})$ $(8b^2)(2b^-8)(2b^-6)$ ?

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How do you simplify $(8 b^{2}) (2 b^{- 8}) (2 b^{- 6})$ $(8b^2)(2b^-8)(2b^-6)$ ?

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Answer 1 · 2016-02-16T12:45:52

$(8 b^{2}) (2 b^{- 8}) (2 b^{- 6}) = 32 b^{- 12}$ $(8b^2)(2b^(-8))(2b^(-6))=color(green)(32b^(-12)$ or $\frac{32}{b^{12}}$ $color(green)(32/b^12)$

Explanation:

Remembering that $b^{p} \times b^{q} = b^{p + q}$ $b^pxxb^q = b^(p+q)$

Re-arrange the factors, grouping the constants together and the powers of $b$ $b$ separately:
$(8 b^{2}) (2 b^{- 8}) (2 b^{- 6})$ $(8b^2)(2b^(-8))(2b^(-6))$
$XXX = (8 \times 2 \times 2) (b^{2} \times b^{- 8} \times b^{- 6})$ $color(white)("XXX")=(8xx2xx2)(b^2xxb^(-8)xxb^(-6))$

$XXX = 32 b^{- 12}$ $color(white)("XXX")=32 b^(-12)$

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How do you simplify $(8 b^{2}) (2 b^{- 8}) (2 b^{- 6})$ $(8b^2)(2b^-8)(2b^-6)$ ?

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