The proof of our result is based on the following series of auxiliary lemmas.
Lemma 1
Let f be a generalized Jordan triple derivation from a ring R to an R bimodule M associated with a Hochschild 2–cocycle map α from R×R into M. Then for all x,y,z∈R, f(xyz+zyx)=f(x)yz+xf(y)z+α(x,y)z+zyf(z)+α(xy,z)+f(z)yx+zf(y)x+α(z,y)x+zyf(x)+α(zy,x).
Proof
Let v=f((x+z)y(x+z)), we have for all x,y,z∈R
$${{} \begin{aligned} 0&=vv\\ &=f(xyx)+f(xyz+zyx)+f(zyz) \{f(x+z)y(x+z)\\ &\,\,\,\,\, +(x+z)f(y)(x+z)+\alpha(x+z,y)(x+z)\\& \quad+(x+z)yf(x+z)\\ &\,\,\,\,\, +\alpha((x+z)y,(x+z))\}. \end{aligned}} $$
Then,
$${{} \begin{aligned} 0&=f(xyx)+f(xyz+zyx)+f(zyz)\{f(x)yx\\&\quad+f(x)yz+f(z)yx\\ &+f(z)yz+xf(y)x+xf(y)z+zf(y)x+zf(y)z\\&\quad+\alpha(x,y)x+\alpha(x,y)z\\ &+\alpha(z,y)x+\alpha(z,y)z+xyf(x)+xyf(z)+zyf(x)\\&\quad+zyf(z)+\alpha(xy,x)\\ &+\alpha(xy,z)+\alpha(zy,x)+\alpha(zy,z)\} \mathrm{\,\,\,for \,\,\, all\,\,\,} x,y,z\in R, \end{aligned}} $$
Therefore,
$${{} \begin{aligned} f(xyz+zyx)&=f(x)yz+xf(y)z+\alpha(x,y)z+zyf(z)\\&\quad+\alpha(xy,z)\\ &\,\,\,\, +f(z)yx+zf(y)x+\alpha(z,y)x+zyf(x)\\ &\,\,\,\, +\alpha(zy,x) \mathrm{\,\,\,for\,\,\, all\,\,\,} x,y,z\in R, \end{aligned}} $$
as required. □
For a generalized Jordan triple derivation f from a ring R to an Rbimodule M associated with a Hochschild 2−cocycle α, we denote by δ, F and β the maps from R×R×R into M defined by δ(x,y,z)=f(xyz)−f(x)yzxf(y)z−α(x,y)zxyf(z)−α(xy,z), F(x,y,z)=f(xyz)−f(x)yzxf(y)zxyf(z) and β(x,y,z)=xyzzyx, respectively. Thus, δ(x,y,z)=F(x,y,z)−α(x,y)z−α(xy,z).
Lemma 2
For all x,y,z in a ring R, the following hold:

(i)
δ(x,y,z)=−δ(z,y,x), and

(ii)
δ(x,y,z) and β(x,y,z) are triadditive.
Proof
(i) Follows easily from Lemma 1.
(ii) Replace x by a+b in the definition of δ, then (ii) is easily seen. □
Lemma 3
For any ring R and any a,b,c,x∈R,
δ(a,b,c)xβ(a,b,c)+β(a,b,c)xδ(a,b,c)=0.
Proof
Let v=f(abcxcba+cbaxabc), then 0=v−v=f((abc)x(cba)+(cba)x(abc))−f(a(bcxcb)a+c(baxab)c). By the definition of the generalized Jordan triple derivation f associated with a Hochschild 2cocycle α and by Lemma 1, we get
$$ {\begin{aligned} 0&=f(abc)xcba+abcf(x)cba+\alpha(abc,x)cba+abcxf(cba)\\ &+\alpha(abcx,cba)+f(cba)xabc+cbaf(x)abc+\alpha(cba,x)abc\\ &+cbaxf(abc)+\alpha(cbax,abc)\{f(a)bcxcba+af(b)cxcba\\ &+abf(c)xcba+abcf(x)cba+ab\alpha(c,x)cba+abcxf(c)ba\\ &+ab\alpha(cx,c)ba+a\alpha(b,cxc)ba+abcxcf(b)a+a\alpha(bcxc,b)a\\ &+\alpha(a,bcxcb)a+abcxcbf(a)+\alpha(abcxcb,a)+f(c)baxabc\\ &+cf(b)axabc+cbf(a)xabc+cbaf(x)abc+cb\alpha(a,x)abc\\ &+cbaxf(a)bc+cb\alpha(ax,a)bc+c\alpha(b,axa)bc+cbaxaf(b)c\\ &+c\alpha(baxa,b)c+\alpha(c,baxab)c+cbaxabf(c)+\alpha(cbaxab,c)\}. \end{aligned}} $$
(1)
Therefore, for all a,b,c,x∈R
$$ {\begin{aligned} 0&=F(a,b,c)xcba+abcxF(c,b,a)\\&+ \{\alpha(abc,x)ab\alpha(c,x)\}cba+ \{\alpha(abcx,cba)\alpha(abcxcb,a)\}\\ &\{ab\alpha(cx,c)ba+a\alpha(b,cxc)ba+ a\alpha(bcxc,b)a+\alpha(a,bcxcb)a\}\\ &+F(c,b,a)xabc+cbaxF(a,b,c)\\ &+\{\alpha(cba,x)cb\alpha(a,x)\}abc+ \{\alpha(cbax,abc)\alpha(cbaxab,c)\}\\ &\{cb\alpha(ax,a)bc+c\alpha(b,axa)bc+ c\alpha(baxa,b)c+\alpha(c,baxab)c\} \end{aligned}} $$
(2)
Since α is a 2–cocycle map, we obtain the following relations for all a,b,c,x∈R:

(i)
{α((ab)c,x)−(ab)α(c,x)}cba={α(ab,cx)−α(ab,c)x}cba.

(ii)
α(abcx,(cb)a)−α((abcx)(cb),a)=α(abcx,cb)a−(abcx)α(cb,a).

(iii)
{α((cb)a,x)−(cb)α(a,x)}abc={α(cb,ax)−α(cb,a)x}abc.

(iv)
α(cbax,(ab)c)−α((cbax)(ab),c)=α(cbax,ab)c−(cbax)α(ab,c).
Substituting from (i–iv) in (2), we get for all a,b,c,x∈R
$$ {\begin{aligned} 0&=F(a,b,c)xcba+abcxF(c,b,a)\\ &+\{\alpha(ab,cx)\alpha(ab,c)x\}cba+ \{\alpha(abcx,cb)aabcx\alpha(cb,a)\}\\ &\{ab\alpha(cx,c)ba+a\alpha(b,cxc)ba+ a\alpha(bcxc,b)a+\alpha(a,bcxcb)a\}\\ &+F(c,b,a)xabc+cbaxF(a,b,c)\\ &+\{\alpha(cb,ax)\alpha(cb,a)x\}abc+ \{\alpha(cbax,ab)ccbax\alpha(ab,c)\}\\ &\{cb\alpha(ax,a)bc+c\alpha(b,axa)bc+ c\alpha(baxa,b)c+\alpha(c,baxab)c\} \end{aligned}} $$
(3)
Since α is a 2–cocycle map, we conclude for all a,b,c,x∈R that
(i) α(ab,cx)=aα(b,cx)+α(a,b(cx))−α(a,b)(cx).
(ii) α(abcx,cb)a={−(abcx)α(c,b)+α((abcx)c,b)+α(abcx,c)b}a.
(iii) α(cb,ax)=cα(b,ax)+α(c,b(ax))−α(c,b)(ax).
(iv) α(cbax,ab)c={−(cbax)α(a,b)+α((cbax)a,b)+α(cbax,a)b}c.
Substituting from (i–iv) in (3), we obtain
$$ {\begin{aligned} 0&=\{F(a,b,c)\alpha(ab,c)\alpha(a,b)c\}xcba+abcx\{F(c,b,a)\alpha(cb,a)\\ &\alpha(c,b)a\}+\{a\alpha(b,cx)cbaab\alpha(cx,c)baa\alpha(b,cxc)ba\}\\ &+\{\alpha(abcxc,b)aa\alpha(bcxc,b)a\alpha(a,bcxcb)a\}\\ &+\alpha(a,bcx)cba+\alpha(abcx,c)ba+\{F(c,b,a)\alpha(cb,a)\\ &\alpha(c,b)a\}xabc+cbax\{F(a,b,c)\alpha(ab,c)\alpha(a,b)c\}\\ &+\{c\alpha(b,ax)abccb\alpha(ax,a)bcc\alpha(b,axa)bc\}\\ &+\{\alpha(cbaxa,b)cc\alpha(baxa,b)c\alpha(c,baxab)c\}\\ &+\alpha(c,bax)abc+\alpha(cbax,a)bc, \,\, \mathrm{for\,\, all}\,\ a,b,c,x\in R. \end{aligned}} $$
(4)
Again since α is a 2–cocycle map, we have

(i)
a{α(b,cx)c−bα(cx,c)−α(b,(cx)c)}ba=−aα(b(cx),c)ba.

(ii)
{α(a(bcxc),b)−aα(bcxc,b)−α(a,(bcxc)b)}a=−α(a,bcxc)ba.

(iii)
c{α(b,ax)a−bα(ax,a)−α(b,(ax)a)}bc=−cα(b(ax),a)bc.

(iv)
{α(c(baxa),b)−cα(baxa,b)−α(c,(baxa)b)}c=−α(c,baxa)bc.
Replacing (i–iv) into (4), we get, for all a,b,c,x∈R
$$ {\begin{aligned} 0&=\delta(a,b,c)xcba+abcx\delta(c,b,a)a\alpha(bcx,c)ba\alpha(a,bcxc)ba\\ &+\alpha(a,bcx)cba+\alpha(abcx,c)ba+\delta(c,b,a)xabc+cbax\delta(a,b,c)\\ &c\alpha(bax,a)bc\alpha(c,baxa)bc+\alpha(c,bax)abc+\alpha(cbax,a)bc. \end{aligned}} $$
(5)
Continuing in this manner, we obtain

(i)
{−aα(bcx,c)−α(a,(bcx)c)+α(a,bcx)c+α(a(bcx),c)}ba=0.

(ii)
{−cα(bax,a)−α(c,(bax)a)+α(c,bax)a+α(c(bax),a)}bc=0.
By (5), we conclude that 0=δ(a,b,c)xcba+abcxδ(c,b,a)+δ(c,b,a)xabc+cbaxδ(a,b,c) for all a,b,c,x∈R. By Lemma 2, we obtain 0=δ(a,b,c)xcba−abcxδ(a,b,c)−δ(a,b,c)xabc+cbaxδ(a,b,c) for all a,b,c,x∈R.
Therefore, δ(a,b,c)xβ(a,b,c)+β(a,b,c)xδ(a,b,c)=0 for all a,b,c,x∈R. This finishes the proof of the lemma. □
Lemma 4
If R is a prime ring of characteristic not 2, then for all a,b,c,x∈R,δ(a,b,c)xβ(a,b,c)=0,.
Proof
By Lemma 3 and Lemma 1.1 of Bre\(\breve {s}\)ar [3], we get the proof. □
Lemma 5
If R is a prime ring of characteristic not 2, then
δ(a_{1},b_{1},c_{1})xβ(a_{2},b_{2},c_{2})=0 for all a_{1},b_{1},c_{1},a_{2},b_{2},c_{2},x∈R.
Proof
From Lemma 2(ii), Lemma 4, and Lemma 1.2 of Bre\(\breve {s}\)ar [3], we get the proof. □
Lemma 6
Let R be a prime ring. Then, R is commutative iff β(a,b,c)=0 for all a,b,c∈R.
Proof
If R is commutative, then, by definition of β,β(a,b,c)=0 for all a,b,c∈R. Conversely, assume that β(a,b,c)=0 for all a,b,c∈R. Let Q be the Martindale right ring of quotients of R defined by Martindale [11]. Then Q is a prime ring with identity that contains the ring R. By Chuang [12], Q satisfies the same generalized polynomial identities as R. In particular abc−cba=0 for all a,b,c∈Q. Replacing c by the identity of Q yields the commutativity of Q, and hence R. □
Lemma 7
Let R be a prime ring of characteristic not 2. Then δ(a,b,c)=0 for all a,b,c∈R, in each of the following cases:

(i)
R is noncommutative.

ii
There exist x,y,z∈R such that β(x,y,z) is a nonzero divisor in M.

iii
R is commutative and α is symmetric.
Proof
(i) By Lemmas 5 and 6, we get our requirement.
(ii) By Lemma 5, we have δ(a,b,c)rβ(x,y,z)=0 for all a,b,c,r,x,y,z∈R. From our assumption δ(a,b,c)r=0 for all a,b,c,r∈R. Thus the primeness of R gives δ(a,b,c)=0 for all a,b,c∈R.
(iii) From Lemma 1 we have f(abc+cba)=f(a)bc+af(b)c+α(a,b)c+abf(c)+α(ab,c)+f(c)ba+cf(b)a+α(c,b)a+cbf(a)+α(cb,a) for all a,b,c∈R. Since R is commutative and α is symmetric, we get 0=2{f(abc)−f(a)bc−af(b)c−abf(c)}−α(a,b)c−α(ab,c)−aα(b,c)−α(a,bc) for all a,b,c∈R. Since α is 2–cocycle we have −aα(b,c)−α(a,bc)=−α(a,b)c−α(ab,c) for all a,b,c∈R. Therefore 0=2{f(abc)−f(a)bc−af(b)c−abf(c)−α(a,b)c−α(ab,c)} for all a,b,c∈R. Since R has characteristic not 2, then δ(a,b,c)=0 for all a,b,c∈R, as required. □