A string is called a square (resp. cube) if it is in the form of XX = X-2 (resp. XXX = X-3). Given a sequence S of length n, a fundamental problem studied in the literature is the problem of computing a longest subsequence of S which is a square or cube (i.e., the longest square/cubic subsequence problem). While the longest square subsequence (LSS) can be computed in O(n(2)) time, the longest cubic subsequence (LCubS) is only known to be solvable in O(n(5)) time, using the longest common subsequence of three strings (LCS-3) as a subroutine (which was much less studied compared with LCS for two strings, or LCS-2). To improve the running time for LCubS, we look at its complementary version and also investigate LCS-3 for three strings S-1, S-2, S-3, with input lengths m <= n(1) <= n(2) respectively. Firstly, we generalize an algorithm by Nakatsu et al. for LCS-2 to have an O(n(1)n(2)delta) algorithm for computing LCS-3, where delta is the minimum number of letters to be deleted in S-1 to have an LCS-3 solution for S-1, S-2 and S-3. This results in an O(k(3)n(2)) algorithm for LCubS, where k is the minimum number of letters deleted in S to have a feasible solution. Then, let R be the number of triples (i, j, k) that match in the input, i.e., S-1[i] = S-2[j] = S-3[k], we show that LCS-3 can be computed in O(n+ R log log n + R-2) time (n is the maximum length of the three input strings). Finally, we define the t-pseudo-subsequence of S under an integer parameter t, which is a string Z containing a subsequence S ' of S such that S ' can be obtained from Z by deleting at most t letters. Subsequently, we study the longest majority t-pseudo-subsequence (LMtPS) of S-i, i = 1..3, which is a t-pseudo-subsequence T = t(1)t(2) center dot center dot center dot t(K) of S-i, i = 1..3, with the maximum length K; moreover, when T is aligned with some subsequence S-i ''s of length K in S-i, i = 1..3, each t(j) matches at least two letters with S-i ', i = 1..3. We show that LMtPS of three strings S-1, S-2 and S-3 is polynomially solvable, while if we require additionally that all letters in Sigma appear in the solution T then it becomes NP-complete, via a reduction to a new SAT instance called Even-(3,B2)-SAT.