Please help meeeeeeee ASAP
1 answer:
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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
If 
is an integer, you can use induction. First show the inequality holds for 
. You have 
, which is true.
Now assume this holds in general for
, i.e. that 
. We want to prove the statement then must hold for 
.
Because, you have
and this must be greater than
for the statement to be true, so we require
for
. Well this is obviously true, because solving the inequality gives 
. So you're done.
If you is any real number, you can use derivatives to show that 
increases monotonically and faster than .
Now assume this holds in general for
Because
and this must be greater than
for
If you