Answer
32$j^{35}$$k^{11}$
Work Step by Step
You are given 4$j^{2}$$k^{6}$(2$j^{11}$$)^{3}$$k^{5}$.Simplify:
4$j^{2}$$k^{6}$2$^3$($j^{11{\times}3}$)$k^{5}$ -Raise the terms in the parentheses to 3-
4$j^{2}$$k^{6}$2$^3$($j^{33}$)$k^{5}$ -add the exponents of like terms since you are multiplying -
4$\times$$2^{3}$$j^{2+33}$$k^{5+6}$
The simplified expression is 32$j^{35}$$k^{11}$
