sum of 2 perfect cubes
a³+b³=(a+b)(x²-xy+y²)
so
x³+4³=(x+4)(x²-4x+16)
set each to zero
x+4=0
x=-4
the other one can't be solveed using conventional means
use quadratic formula
for
ax^2+bx+c=0
x=[tex] \frac{-b+/- \sqrt{b^2-4ac} }{2a} [/tex]
for x²-4x+16=0
x=[tex] \frac{-(-4)+/- \sqrt{(-4)^2-4(1)(16)} }{2(1)} [/tex]
x=[tex] \frac{4+/- \sqrt{16-64} }{2} [/tex]
x=[tex] \frac{4+/- \sqrt{-48} }{2} [/tex]
x=[tex] \frac{4+/- (\sqrt{-1})(\sqrt{48}) }{2} [/tex]
x=[tex] \frac{4+/- (i)(4\sqrt{3}) }{2} [/tex]
x=[tex] 2+/- 2i\sqrt{3} [/tex]
the roots are
x=-4 and 2+2i√3 and 2-2i√3
