How do you simplify ${(\frac{a^{- 2} b^{4} c^{5}}{a^{- 4} b^{- 4} c^{3}})}^{2}$ $((a^-2b^4c^5)/(a^-4b^-4c^3))^2$ ?

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How do you simplify ${(\frac{a^{- 2} b^{4} c^{5}}{a^{- 4} b^{- 4} c^{3}})}^{2}$ $((a^-2b^4c^5)/(a^-4b^-4c^3))^2$ ?

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Answer 1 · 2017-07-14T17:02:45

${(\frac{a^{- 2} b^{4} c^{5}}{a^{- 4} b^{- 4} c^{3}})}^{2} = a^{4} b^{16} c^{4}$ $((a^(-2)b^4c^5)/(a^(-4)b^(-4)c^3))^2=a^4b^16c^4$

Explanation:

Let us use the identities $a^{- m} = \frac{1}{a^{m}}$ $a^(-m)=1/a^m$ , $\frac{1}{a^{- m}} = a^{m}$ $1/a^(-m)=a^m$ , $a^{m} \times a^{n} = a^{m + n}$ $a^mxxa^n=a^(m+n)$ , $\frac{a^{m}}{a^{n}} = a^{m - n}$ $a^m/a^n=a^(m-n)$ and ${(a^{m})}^{n} = a^{m n}$ $(a^m)^n=a^(mn)$

Hence ${(\frac{a^{- 2} b^{4} c^{5}}{a^{- 4} b^{- 4} c^{3}})}^{2}$ $((a^(-2)b^4c^5)/(a^(-4)b^(-4)c^3))^2$

= ${(\frac{\frac{1}{a^{2}} b^{4} c^{5}}{\frac{1}{a^{4}} \times \frac{1}{b^{4}} c^{3}})}^{2}$ $((1/a^2b^4c^5)/(1/a^4xx1/b^4c^3))^2$

= ${(\frac{a^{4} b^{4} b^{4} c^{5}}{a^{2} c^{3}})}^{2}$ $((a^4b^4b^4c^5)/(a^2c^3))^2$

= ${(a^{4 - 2} b^{4 + 4} c^{5 - 3})}^{2}$ $(a^(4-2)b^(4+4)c^(5-3))^2$

= ${(a^{2} b^{8} c^{2})}^{2}$ $(a^2b^8c^2)^2$

= $a^{2 \times 2} b^{8 \times 2} c^{2 \times 2}$ $a^(2xx2)b^(8xx2)c^(2xx2)$

= $a^{4} b^{16} c^{4}$ $a^4b^16c^4$

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How do you simplify ${(\frac{a^{- 2} b^{4} c^{5}}{a^{- 4} b^{- 4} c^{3}})}^{2}$ $((a^-2b^4c^5)/(a^-4b^-4c^3))^2$ ?

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Explanation:

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How do you simplify (a−2b4c5a−4b−4c3)2((a^-2b^4c^5)/(a^-4b^-4c^3))^2?

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How do you simplify ${(\frac{a^{- 2} b^{4} c^{5}}{a^{- 4} b^{- 4} c^{3}})}^{2}$ $((a^-2b^4c^5)/(a^-4b^-4c^3))^2$ ?