Is this possible?

It is given that: $\begin{cases} a + b = 4 & ...(1) \\ a^{\frac{1}{3}} + b = 5 & ...(2) \\ a + b^{\frac{1}{3}} = 6 & ...(3) \end{cases}$

$\begin{aligned} (2)+(3)-(1): \quad a^{\frac{1}{3}} + b^{\frac{1}{3}} & = 7 \quad ...(4) \\ a^{\frac{2}{3}} + b^{\frac{2}{3}} & = x \quad ...(5) \\ \Rightarrow \left(\color{#3D99F6}{a^{\frac{1}{3}} + b^{\frac{1}{3}}}\right)^2 - 2 \left(ab\right)^{\frac{1}{3}} & = x \\ \color{#3D99F6}{7}^2 - 2 \left(ab\right)^{\frac{1}{3}} & = x \quad \quad \small \color{#3D99F6}{(4)} \\ \Rightarrow \left(ab\right)^{\frac{1}{3}} & = \frac{49-x}{2} \quad ...(6) \end{aligned}$

$\begin{aligned} (1): \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad a + b & = 4 \\ \left(\color{#3D99F6}{a^{\frac{1}{3}} + b^{\frac{1}{3}}}\right) \left(\color{#D61F06}{a^{\frac{2}{3}} + b^{\frac{2}{3}}} - \color{#20A900}{\left(ab\right)^{\frac{1}{3}}}\right) & = 4 \\ \color{#3D99F6}{7} \left(\color{#D61F06}{x} - \color{#20A900}{\frac{49-x}{2}}\right) & = 4 \quad \quad \small \color{#3D99F6}{(4)}, \space \color{#D61F06}{(5)}, \space \color{#20A900}{(6)} \\ 14x - 343 + 7x & = 8 \\ \Rightarrow x & = \frac{351}{21} = 16.71428571 \\ \Rightarrow \lfloor 1000x \rfloor & = \boxed{16714} \end{aligned}$

From the given equations, $\sqrt[3]{a} + \sqrt[3]{b} = 7.$ Cubing this equation, we get $a + 3 \sqrt[3]{a^2 b} + 3 \sqrt[3]{ab^2} + b = 343.$ We know that $a + b = 4$ and $3 \sqrt[3]{a^2 b} + 3 \sqrt[3]{ab^2} = 3 \sqrt[3]{ab} (\sqrt[3]{a} + \sqrt[3]{b}) = 21 \sqrt[3]{ab}$ , so $\sqrt[3]{ab} = \frac{343 - 4}{21} = \frac{113}{7}.$

Squaring the equation $\sqrt[3]{a} + \sqrt[3]{b} = 7$ , we get $\sqrt[3]{a^2} + 2 \sqrt[3]{ab} + \sqrt[3]{b^2} = 49,$ so $\sqrt[3]{a^2} + \sqrt[3]{b^2} = 49 - 2 \cdot \frac{113}{7} = \frac{117}{7}.$ This appears to give an answer to the problem, but there is more to the story.

Squaring this equation, we get $a \sqrt[3]{a} + 2 \sqrt[3]{a^2 b^2} + b \sqrt[3]{b} = \frac{13689}{49},$ so $a \sqrt[3]{a} + b \sqrt[3]{b} = \frac{13689}{49}- 2 \left( \frac{113}{7} \right)^2 = -\frac{11849}{49}.$

But multiplying the equations $\sqrt[3]{a} + b = 5$ and $a + \sqrt[3]{b} = 6$ , we get $(\sqrt[3]{a} + b)(a + \sqrt[3]{b}) = 30,$ so $a \sqrt[3]{a} + b \sqrt[3]{b} + ab + \sqrt[3]{ab} = 30.$ Hence, $a \sqrt[3]{a} + b \sqrt[3]{b} = 30 - \frac{113}{7} - \left( \frac{113}{7} \right)^3 = -\frac{1438144}{343}.$ This value is different from above, so the given system has no solutions, and the "answer" to the problem is 567.

$\left( \sqrt [ 3 ]{ a } +b \right) +\left( a+\sqrt [ 3 ]{ b } \right) -\left( a+b \right) =7$

$\sqrt [ 3 ]{ a } +\sqrt [ 3 ]{ b } =7$

$a+b=\left( \sqrt [ 3 ]{ a } +\sqrt [ 3 ]{ b } \right) \left( { a }^{ \frac { 2 }{ 3 } }+{ b }^{ \frac { 2 }{ 3 } }+\sqrt [ 3 ]{ ab } \right)$

$4=7\left( x+\sqrt { ab } \right)$

${ \left( \sqrt [ 3 ]{ a } +\sqrt [ 3 ]{ b } \right) }^{ 2 }=\left( { a }^{ \frac { 2 }{ 3 } }+{ b }^{ \frac { 2 }{ 3 } }+\sqrt { ab } \right)$

$49=\left( x+2\sqrt { ab } \right)$

$\frac { 4 }{ 7 } =x-\sqrt { ab }$

$\frac { 8 }{ 7 } =2x-2\sqrt { ab }$

$\frac { 8 }{ 7 } +49=2x-2\sqrt { ab } +\left( x+2\sqrt { ab } \right)$

$\frac { 351 }{ 7 } =3x$

$\frac { 117 }{ 7 } =x$

so enter 16714

OR

$sub\quad m=\sqrt [ 3 ]{ a } ;sub\quad n=\sqrt [ 3 ]{ b } \\ then\\ { m }^{ 3 }+{ n }^{ 3 }=4\\ m+{ n }^{ 3 }=5\\ { m }^{ 3 }+{ n }=6\\ { m }^{ 2 }+{ n }^{ 2 }=x\\ { m }^{ 3 }+{ n }^{ 3 }+m+n=11\\ m+n=7\\ { \left( m+n \right) }^{ 3 }={ 7 }^{ 3 }\\ { m }^{ 3 }+{ n }^{ 3 }+3mn(m+n)=343\\ 4+3mn(7)=343\\ mn=\frac { 113 }{ 7 } \\ since\quad x={ m }^{ 2 }+{ n }^{ 2 }\\ \qquad \qquad ={ (m+n) }^{ 2 }-2mn\\ \qquad \qquad ={ (7) }^{ 2 }-2\left( \frac { 113 }{ 7 } \right) \\ \qquad \qquad =\frac { 117 }{ 7 } \\ hence,\\ \left\lfloor 1000x \right\rfloor =16714$