How would one do the asymptotic analysis of Strassen's algorithm to show $T(n)$ is $Θ(n^{log_27})$?
I know $T(n) = 7T(n/2) = 7^2T(n/4) = ... 7^{log_2n} + 1, T(n) = cd^4$ where $d = log_ba $ and $ c = T(1)$ a constant, and $7^{log_2n} = 7^{log_7n * log_27} = (7^{log_7n})^{log_27} = n^{log_27}$ is this sufficient?
Why are you squaring 7^2 and where does the cd^4 come from? Approach this from a difference angle. Think about the simple recurrence solution to T(n) = 7T(n/2). Then, you expect to find a solution of the form c times n^(log base 2 of 7) + \theta n^2, therefore making your substitute for T(n) as c_1 times n^(log base 2 of 7) + n^2. The rest should follow once you set this substitution for T(n) equal to the original Strassen's algorithm, 7T(n/2) + n^2