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=
for x²-4x+16=0
x=
x=
x=
x=
x=
x=
the roots are
x=-4 and 2+2i√3 and 2-2i√3
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=
for x²-4x+16=0
x=
x=
x=
x=
x=
x=
the roots are
x=-4 and 2+2i√3 and 2-2i√3
