How do you simplify ${(3 n^{2} m^{4})}^{2}$ $(3n^2m^4)^2$ ?

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Answer 1 · 2015-10-30T11:41:10

$9 n^{4} m^{8}$ $9n^4m^8$

By using laws of exponents we get

${(3 n^{2} m^{4})}^{2} = 3^{2} \cdot {(n^{2})}^{2} \cdot {(m^{4})}^{2}$ $(3n^2m^4)^2=3^2*(n^2)^2*(m^4)^2$

$= 9 n^{4} m^{8}$ $=9n^4m^8$

How do you simplify (3n2m4)2(3n^2m^4)^2?