Simplify ${(a^{3} b^{2})}^{- 2} \times {(a^{2} b^{3})}^{4}$ $(a^3b^2)^-2 xx (a^2b^3)^4$ ?

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Answer 1 · 2016-08-16T02:29:27

$a^{2} b^{8}$ $a^2b^8$

Apply two of the rules of indices:

${(a^{n})}^{m} = a^{n \times m}$ $(a^n)^m = a^(nxxm)$ (1)

$a^{n} \times a^{m} = a^{n \times m}$ $a^nxxa^m = a^(nxxm)$ (2)

Here our expression is: ${(a^{3} b^{2})}^{- 2} \times {(a^{2} b^{3})}^{4}$ $(a^3b^2)^-2 xx (a^2b^3)^4$

Apply (1) to both terms: $\to a^{- 6} b^{- 4} \times a^{8} b^{12}$ $-> a^(-6)b^(-4) xx a^8b^12$

Group powers of $a$ $a$ & $b$ $b$ and apply (2) $\to a^{(- 6 + 8)} \times b^{(- 4 + 12)}$ $-> a^((-6+8))xxb^((-4+12))$

$= a^{2} \times b^{8} = a^{2} b^{8}$ $= a^2xxb^8 = a^2b^8$

Simplify (a3b2)−2×(a2b3)4(a^3b^2)^-2 xx (a^2b^3)^4 ?