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
Answer:
Step-by-step explanation:
x+ 8/4 -x+ 5/4
= x+2+-x +5/4
Combine Like Terms:
x+2+-x+ 5/4
(x+x)+ ( 2 +5/4)
= 13/4
B=48/7 you simplify 7/b(b) to 7b/6
Then multiply both sides by 6 so 8 x 6 =7b then simplify 8 x 6 =48
And divide both sides by 7 48\7 then switch sides b=48/7
x+3<8
8-3 =5
X<5
any number less than 5 will work