Try Gonometry

Sum of Two Cubes: $\sin^3 x + \cos^3 x = (\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)$ First, note that from the first equation you get: $(\sin x + \cos x)^2 = \dfrac{1}{2}^2$ $1+2\sin x\cos x = \dfrac{1}{4}$ $\sin x\cos x = -\dfrac{3}{8}$ Now we have everything we need, we just need to substitute these values to the first equation: $(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)$ $\left(\dfrac{1}{2}\right)\left(1+\dfrac{3}{8}\right)$ $= \boxed{\dfrac{11}{16}}$ And so, $b-a=16-11=\boxed{5}$

Given:

$\sin { (x)\quad +\quad } \cos { (x) } \quad =\quad \frac { 1 }{ 2 }$

Square both sides

${ \left[ \sin { (x) } \quad +\quad \cos { (x) } \right] }^{ 2 }\quad =\quad { \left( \frac { 1 }{ 2 } \right) }^{ 2 }$

Factor out

$\sin ^{ 2 }{ (x)\quad +\quad 2\sin { (x) } \cos { (x)\quad +\quad \cos ^{ 2 }{ (x)\quad =\quad \frac { 1 }{ 4 } } } }$

But

$\sin ^{ 2 }{ (x)\quad +\quad } \cos ^{ 2 }{ (x) } \quad =\quad 1$

So

$2\sin { (x) } \cos { (x)\quad +\quad 1\quad =\quad \frac { 1 }{ 4 } }$

Simplify (Eq. 1)

$\sin { (x) } \cos { (x) } \quad =\quad -\frac { 3 }{ 8 } \quad Eq.\quad (1)$

Cube given equation

${ \left[ \sin { (x)\quad +\quad \cos { (x) } } \right] }^{ 3 }\quad =\quad { \left( \frac { 1 }{ 2 } \right) }^{ 3 }$

Factor out

$\sin ^{ 3 }{ (x) } \quad +\quad 3\sin ^{ 2 }{ (x) } \cos { (x) } \quad +\quad 3\sin { (x) } \cos ^{ 2 }{ (x) } \quad +\quad \cos ^{ 3 }{ (x) } \quad =\quad \frac { 1 }{ 8 }$

Separate

$\sin ^{ 3 }{ (x) } \quad +\quad \cos ^{ 3 }{ (x) } \quad =\quad \frac { 1 }{ 8 } \quad -\quad 3\sin ^{ 2 }{ (x) } \cos { (x) } \quad -\quad 3\sin { (x) } \cos ^{ 2 }{ (x) }$

Factor out right side

$\sin ^{ 3 }{ (x) } \quad +\quad \cos ^{ 3 }{ (x) } \quad =\quad \frac { 1 }{ 8 } \quad -\quad 3\sin { (x) } \cos { (x) } \left[ \sin { (x) } \quad +\quad \cos { (x) } \right]$

Plugin given equation and Eq. 1

$\sin ^{ 3 }{ (x) } \quad +\quad \cos ^{ 3 }{ (x) } \quad =\quad \frac { 1 }{ 8 } \quad -\quad 3\left( -\frac { 3 }{ 8 } \right) \left( \frac { 1 }{ 2 } \right) \quad =\quad \frac { 11 }{ 16 } \quad =\quad \frac { a }{ b }$

$\therefore \quad a\quad =\quad 11\quad and\quad b\quad =\quad 16$

$b\quad -\quad a\quad =\quad 16\quad -\quad 11\quad =\quad \boxed { 5 }$