Tom, you don't know what the value of "a" is....
SO, dividing both sides of the equation
16a[sup:14hawj76]3[/sup:14hawj76] = 4a
by "a" is a VERY RISKY proposition. What if a = 0?? Is it OK to divide by 0? I think not.
Here's a much better approach.
You've got
16a[sup:14hawj76]3[/sup:14hawj76] = 4a
Subtract 4a from both sides of the equation:
16a[sup:14hawj76]3[/sup:14hawj76] - 4a = 4a - 4a
16a[sup:14hawj76]3[/sup:14hawj76] - 4a = 0
Now, factor the expression on the left side. Start by removing a common factor of 4a:
4a (4a[sup:14hawj76]2[/sup:14hawj76] - 1) = 0
Now look at the expression inside the parentheses. Do you see that 4a[sup:14hawj76]2[/sup:14hawj76] - 1 is a difference of two squares? 4a[sup:14hawj76]2[/sup:14hawj76] is (2a)[sup:14hawj76]2[/sup:14hawj76] - 1[sup:14hawj76]2[/sup:14hawj76]
So, (2a)[sup:14hawj76]2[/sup:14hawj76] - 1[sup:14hawj76]2[/sup:14hawj76] factors as (2a + 1)(2a - 1)
And your factored expression is
4a (2a + 1)(2a - 1) = 0
If the product of two or more factors is 0, then at least ONE of those factors must have a value of 0. So, if
4a (2a + 1) (2a - 1) = 0
then
4a = 0
OR
2a + 1 = 0
OR
2a - 1 = 0
Solve each of those simple equations. You'll end up with THREE solutions.