Factoring Cubes

We know that $\large(x+y)^3=x^3+3x^2y+3xy^2+y^3$ . We apply this to $\large x+\dfrac{1}{x}=4$ .

$\large \left(x+\dfrac{1}{x}=4\right)^3$ $\implies$ $\large x^3+3x+\dfrac{3}{x}+\dfrac{1}{x^3}=64$ $\implies$ $\large x^3+\dfrac{1}{x^3}=64-3x-\dfrac{3}{x}$ $\implies$ $\large x^3+\dfrac{1}{x^3}=64-3\left(x+\dfrac{1}{x}\right)$

$\implies$ $\large x^3+\dfrac{1}{x^3}=64-3(4)=64-12=$ $\large \color{#D61F06}\boxed{52}$

You can cube both sides, providing

$(x+\dfrac{1}{x})^3=4^3$

$x^3+\dfrac{1}{x^3}+3x+\dfrac{3}{x}=64$

$x^3+\dfrac{1}{x^3}+3(x+\dfrac{1}{x})=64$

$x^3+\dfrac{1}{x^3}+3(4)=64$

$x^3+\dfrac{1}{x^3}=64-12=\color{#20A900}\large\boxed{52}$

$x+\frac { 1 }{ x } =4\\ { \left( x+\frac { 1 }{ x } \right) }^{ 3 }={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } +3{ x }^{ 2 }\frac { 1 }{ x } +3x\frac { 1 }{ { x }^{ 2 } } \\ { 4 }^{ 3 }={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } +3x\frac { 1 }{ x } \left( x+\frac { 1 }{ x } \right) \\ 64={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } +3\cdot 4\\ 52={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } }$

A well known cubic identity is $a^3+b^3=(a+b)(a^2-ab+b^2)$

We can apply this to $x^3+\frac{1}{x^3}$ as follows

$x^3+\frac{1}{x^3}=x^3+(\frac{1}{x})^{3}=(x+\frac{1}{x})(x^2+(\frac{1}{x})^2-1)$

Notice $x^2+(\frac{1}{x})^2=(x+\frac{1}{x})^2-2$

So $x^3+\frac{1}{x^3}=(x+\frac{1}{x})((x+\frac{1}{x})^2-2-1)$

$x^3+\frac{1}{x^3}=(4)((4)^2-3)$

$x^3+\frac{1}{x^3}=52$ , and we have our answer

a^3+b^3=(a+b)^3 - 3ab(a+b)

so,x^3+1/x^3=(4)^3-3 x 4=64-12=52