Cubic summation 2

$\sum _{ x=1 }^{ \infty } \frac { { 3x }^{ 2 }+12x+16 }{ { (x(x+1)(x+2)(x+3)(x+4)) }^{ 3 } } \\ \quad \\ \text{the above expression can be rewritten as}\\ \\ \sum _{ 1 }^{ \infty }{ \frac { { (x+4) }^{ 3 }-{ (x) }^{ 3 } }{ { (x(x+1)(x+2)(x+3)(x+4)) }^{ 3 } } } *\frac { 1 }{ 4 } \\ \quad \\ which\quad will\quad give\quad us\\ \\ \frac { 1 }{ 4 } \sum _{ 1 }^{ \infty }{ \frac { 1 }{ { (x(x+1)(x+2)(x+3)) }^{ 3 } } } -\frac { 1 }{ { ((x+1)(x+2)(x+3)(x+4)) }^{ 3 } } \\ \\ \\ thus\quad we\quad can\quad see\quad that\quad it\quad would\quad form\quad a\quad telescopic\quad sum\\ where\quad all\quad the\quad terms\quad except\quad the\quad first\quad one\quad will\quad get\quad cancelled\quad out.\\ thus\quad our\quad answer\quad would\quad be\\ \frac { 1 }{ { (1*2*3*4) }^{ 3 } } *\frac { 1 }{ 4\\ } =\quad \frac { 1 }{ { (4!) }^{ 3 } } *\frac { 1 }{ 4 } \\ \\ thus\quad our\quad a\quad would\quad be\quad 4\quad and\quad b\quad would\quad be\quad 3.\\ thus\quad the\quad final\quad answer\quad would\quad be\quad B-A\quad =\quad 3-4\quad =\quad -1.\\ \\ \\ \\ \\ \\$