Answer
$32j^{35}k^{11}$
Work Step by Step
Given expression : $4j^{2}k^{6}(2j^{11})^{3}k^{5}$
By law of exponents, $(ab)^{n} = a^{n}b^{n}$
Hence, $(2j^{11})^{3} = 2^{3}$ X$( j^{11})^{3} = 8$ X $j^{33}$
Therefore, the given expression = $4$ X $j^{2}$ X $k^{6}$ X $8$ X $j^{33}$ X $k^{5}$
This becomes : $32$ X $j^{2+33}$ X $k^{6+5}$
(since the exponents are added while multipliying with the same base)
This becomes : $32$ X $j^{35}$ X $k^{11}$
Final result : $32j^{35}k^{11}$
