How do you simplify $(4 a^{3} b^{3}) {(4 a b^{2})}^{2}$ $(4a^3b^3)(4ab^2)^2$ leaving only positive exponents?

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Answer 1 · 2017-02-28T19:57:05

$64 a^{5} b^{7}$ $64a^5b^7$

Split it down into easier parts.

${(4 a b^{2})}^{2}$ $(4ab^2)^2$

Multiply all the inside indices by $2$ $2$ because of the $2$ $2$ on the outside:

${(4 a b^{2})}^{2} = 4^{2} a^{2} b^{4} = 16 a^{2} b^{4}$ $(4ab^2)^2 = 4^2a^2b^4 = 16a^2b^4$

Now we have

$(4 a^{3} b^{3}) (16 a^{2} b^{4})$ $(4a^3b^3)(16a^2b^4)$

You should know that $a^{b} \cdot a^{c} = a^{b + c}$ $a^b * a^c = a^(b+c)$ , so

$4 \cdot a^{3} \cdot b^{3} \cdot 16 \cdot a^{2} \cdot b^{4}$ $4 * a^3 * b^3 * 16 * a^2 * b^4$

$= 4 \cdot 16 \cdot a^{3 + 2} \cdot b^{3 + 4}$ $= 4 * 16 * a^(3+2) * b^(3+4)$

$= 64 a^{5} b^{7}$ $=64a^5b^7$

How do you simplify (4a3b3)(4ab2)2(4a^3b^3)(4ab^2)^2 leaving only positive exponents?