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
There are usually 16 games in a football season.
Since there's three more games, they've had 13 so far.
There are 27 combinations of outcomes.
Hopefully this helps, and sorry I didn't do 8 I'm limited on time
Hello,
Let's assume n,n+1,n+2,n+3,n+4 the 5 numbers
n+(n+1)+...+(n+4)=5n+10=265
5n=265-10
5n=255
n=51
The 5th number is 51+4=55