What's x in (x+3)^3-(3x-1)^2 = x^3+4?

Expand the terms and simplify:

 

First term:   (x + 3)³ = x³ +9x² + 27x + 27

2nd term:    (3x - 1)² = 9x² - 6x + 1

 

Subtract the second from the first:   x³ + 33x + 26

This must equal: x³ + 4 so

 

 x³ + 33x + 26 = x³ + 4

33x + 26 = 4

33x = -22

x = -22/33 = -2/3

 

Check:

$${\left({\mathtt{\,-\,}}{\frac{{\mathtt{2}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{2}}}{{\mathtt{3}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}^{{\mathtt{2}}} = {\frac{{\mathtt{100}}}{{\mathtt{27}}}} = {\mathtt{3.703\: \!703\: \!703\: \!703\: \!703\: \!7}}$$

$${{\mathtt{\,-\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}} = {\frac{{\mathtt{100}}}{{\mathtt{27}}}} = {\mathtt{3.703\: \!703\: \!703\: \!703\: \!703\: \!7}}$$

 

Ok.