If $x^3 + y^3 = z^3$ then it is easy to prove that without loss of generality a co-prime version must exist such that $x, y, z$ have no common factor other than 1. Further it is easy to prove that there must be two co-prime positive numbers $p$ and $q$ such that one of the three, say $z$, has the form $z^3 = 2p(p^2 + 3q ^2)$ with $p$ and $q$ of opposite parity (one odd, the other even), therefore $p^2 + 3q^2$ being an odd cube and $2p$ being an even cube, and gcd $(2p, p^2 + 3q^2)$ being 1 or 3. As Euler has shown, it is possible then to find a smaller solution to Fermat's problem  for $n$ = 3. The proof of the last step is somewhat [tedious](http://fermatslasttheorem.blogspot.de/2005/05/fermats-last-theorem-proof-for-n3.html). In both cases it includes the lemma that there exist two co-prime numbers $a, b$ of opposite parity such that $p = a^3 - 9ab^2$.

Perhaps Tait thought to use the result of his [equation (a)](http://www.archive.org/stream/proceedingsroya40edingoog#page/n166/mode/1up), "every cube is the difference of two squares, one at least of which is divisible by 9" in order to easen this last step.

Concerning part (b), it is obvious that Tait's second equation has the same structure as the original one, but now containing three cubes that are definitely greater than the original cubes.

$((x^3 + z^3)y)^3 + ((x^3 - y^3)z)^3 = ((z^3+y^3)x)^3$

This procedure of increasing the cubes can be repeated in infinity. Although it is not justified to reverse the derivation and to conclude that by this method every triple of cubes could be reduced to smaller cubes, it cannot be excluded that Tait was misled by this slip.

I don't believe that there is an "easy" proof, and I have to offer three reasons:

The first one (and admittedly the weakest) is that I have not suceeded in finding an easy solution. Any simplification like

$(2x^3 + y^3)^3y^3 + (x^3 - y^3)^3(x^3+y^3) = (x^3 + 2y^3)^3x^3$

$((2x^3 + y^3)^3 + (x^3 - y^3)^3)y^3 = ((2y^3 + x^3)^3 + (y^3 - x^3)^3)x^3$

leads easily to the result that the derived equation is correct, finally leaving 0 = 0, but no information about the impossibility of an integer character of the roots appeares obvious to me.

A certainly stronger reason is that this, if correct very interesting, note has, as far as I can see, never been mentioned again by Tait and has not been included in his [collected papers](http://archive.org/details/scientificpapers01taituoft), which were edited and furnished with a preface by Tait himself. (The note has however been mentioned in a great many of other sources including [Ribenboim's popular book](http://math.sjtu.edu.cn/course/skymath/skymathinfo/referencebook/fermat's_last_theorem.pdf) and [Wikipedia](http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem).) 

The third and most important reason however, outweighing all others, is that this question has been asked in the American Mathematical Monthly in [1914](http://www.jstor.org/stable/2974261), [1916](http://www.jstor.org/stable/2974032), [1919](http://www.jstor.org/stable/2973144), [1920](http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem), [1921](http://www.jstor.org/stable/2972390), and [1922](http://www.jstor.org/stable/2298891?seq=1), which at that time was edited by eminent mathematicians like Hurwitz and certainly read by many others. Last but not least, the problem has been open here in MathOverflow for more than one year. No easy way could be shown.

So it is very probably that this note belongs to the same category as Fermat's original one.