How do you simplify ${(3 a^{4} b^{2})}^{3}$ $(3a^4b^2)^3$ ?

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Answer 1 · 2017-10-24T11:01:53

$27 a^{12} b^{6}$ $27a^12b^6$

Move the $3$ $3$ inside by simply multiplying every power by $3$ $3$ .

${(3 a^{4} b^{2})}^{3} = {(3^{1} a^{4} b^{2})}^{3}$ $(3a^4b^2)^3 = (3^1a^4b^2)^3$

$= (3^{1 \times 3} a^{4 \times 3} b^{2 \times 3})$ $= (3^(1xx3)a^(4xx3)b^(2xx3))$

$= 3^{3} a^{12} b^{6}$ $= 3^3a^12b^6$

You can calculate $3^{3}$ $3^3$ straight away, because

$3^{3} = 3 \times 3 \times 3 = (3 \times 3) \times 3 = 9 \times 3 = 27$ $3^3 = 3xx3xx3 = (3xx3)xx3 = 9xx3 = 27$

so

${(3 a^{4} b^{2})}^{3} = 27 a^{12} b^{6}$ $(3a^4b^2)^3 = 27a^12b^6$

How do you simplify (3a4b2)3(3a^4b^2)^3?