The operation a # b is defined by a + 2b + 3ab. Solve the equation (a # a) # a = a # (a # a).
a # b = a + 2b + 3ab
For (a # a ) # a:
Working inside the parentheses first:
a # a =
The second a takes the place of b in the formula: a # b = a + 2b + 3ab
so we have: a # a = a + 2a + 3aa = 3a + 3a2
Now we have: (3a + 3a2) # a
In this case, the expression "3a + 3a2" takes the place of the a
while a takes the place of b
so we have: a # a = (3a + 3a2) + 2a + 3(3a + 3a2)a
= 3a + 3a2 + 2a + 9a2 + 9a3
= 5a + 13a2 + 9a3
This is the left-hand side. Can you do the right-hand side?
If you can, set the two sides equal to each other and solve ...