This Question is very simple, Only thing we do approaches to the problem in wrong way.

L = 0*(10*)* : To do this type of problem always expand * As precedence of operator is [(*) >( .) > (+)]. Now you will never do this wrong.

hence the L = 0*(10*)* can be expanded like this ->

= 0*{ e + (10*) + (10*)(10*) + (10*)(10*)(10*) +..........}

Now after this just take Alphabet(A) = { 0, 1} and find the what are the strings present in A* or {0,1} * can also be obtain by L = 0*(10*)*.

## You will find that L = {0, 1} * = A *.( Which is also a STANDARD RESULT, if you want, you can remember this)

** Ex - We can generate 100 from L by doing like this -> 0*(10*) = e.(100) = 100.**

SIMILARLY by doing the same way,

M = (1*0)*1*

= {e + (1*0) + (1*0)(1*0) + (1*0)(1*0)(1*0) + ..........} 1*

By doing the same way you can easily find that it is also equal to {0,1}* or A*.

**Ex - We can also generate 100 from M also in this way -> just take (1*0)(1*0) = (10)(1*0) = (10)(0) = 100.**

Hence Both M and L are equal .