#Help pls)
76 messages · Page 1 of 1 (latest)
Hii)
factorize x^3 + y^3
(x+y) (x^2-xy+y^2) = 19
(x+y) (xy + 8) = 2
Yea so)
seems suspicious
?
(x+y) (x^2-xy+y^2 - xy -8) = 17
just keep messign around
not sure if it is the right response but messing around will work
And i get (x+y)(x^2-2xy+y^2-8)=17
So?
And after that, i can move the 8 to the right side
(x+y) (x-y)^2=25
but 8 is in the brackets
Oops
hmm but i m pretty sure that we r close
OH HERE
here i got it
i said minus becuz i was araid (x^2-xy+y^2) and (xy + 8) were equal to 0 so u cant divide
Idk how to solve it xd
but u can proof that they are not equal to 0
that means
(x+y) (x^2-xy+y^2) = 19
(x+y) (xy + 8) = 2
x+ y = 19/ (x^2-xy+y^2)
x+ y = 2/ (xy + 8)
u get it ?
Not so much
fudge
y i keep getting delusional
hmm just leave it there for a sec
ill ping u when i get it
if not im prolyl dead
naur bro i failed u :( (i had to go to sleep 💀 )
text me when ur given the ans
:( ok
I got something::
From you guys' results i found out that: x/2 = (76-y^2)/2x-17y
But if we could cancel out the xy term in some way i think we could solve this one
@hardy vigil
$$19=(x+y)[(x+y)^2-3xy]$$ $$2=(xy+8)(x+y)$$
Civil Service Pigeon
Let $x+y=a, xy=b$: $$19=a(a^2-3b)$$ $$2=a(b+8)$$
Civil Service Pigeon
the rest is decently straightforward
and how did you get the first expression? Could you explain, please?
from x^2 - xy + y^2
wow u didnt sleeped
nice xd
(ye i rlly didnt_
oh really
ye :3
It was hard to think of that
.close