Will someone

[A Former User]

JDP_Random no

[JDP_Random]

… well at least I still got an A+!

[A Former User]

No

[A Former User]

JDP_Random no

[JDP_Random]

Well A B+ is still good!

[A Former User]

JDP_Random you copied your work from tripod

[A Former User]

JDP_Random F 0%

[JDP_Random]

But i showed mine

[A Former User]

No, you may have an F instead

[A Former User]

JDP_Random but you copied from him

[JDP_Random]

but i got 42.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000001

[A Former User]

JDP_Random then F cuz you got the wrong answer

[A Former User]

Plus the wrong process

[JDP_Random]

.

[A Former User]

JDP_Random F

[A Former User]

Nah that could be still interpreted as a 59%, Let’s say 0 instead

[JDP_Random]

This aint fair at all

[A Former User]

JDP_Random shouldn’t have cheated 💀

[MGC]

@I_Løve_Zøf for k=1
x=cube root of A; y=cube root of B; z=cube root of C
where A+B+C=k
A,B,C can be any numbers, for k=1 for example A=0,25; B=0,4; C=0,35

crX^3+crY^3+crZ^3=k
cr(0,25)^3+cr(0,4)^3+cr(0,35)^3=1
0,25+0,4+0,35=1
1=1

For any other k, multiply both sides of that equation by k
k(cr(0,25)^3+cr(0,4)^3+cr(0,35)^3)=k*1
0,25k+0,4k+0,35k=k
k=k

eat shit, the amount of solutions is infinite considering x,y,z can be any number with any amount of numbers after decimal point

[A Former User]

MGC 42