@andymath.com: Can you spot the mistake? It’s tiny #andymath #math #maths

Usman🇵🇰🇦🇹🔻 :

day 2 asking you to do this

pruria :

how exciting

mathiaskanayao :

check the identity, it's wrong

Sean :

Does that make sense y?

Timothy Simington :

can you figure what what the X and Y value can be???

Hazeez :

can I knw the application you're using for this amazing work ???

Hema_657 :

porimol roy :

@ 😏

claud_25 :

I did a substitution, like N, and solved the first for N, to get 7±4√3, then, wrote N√N+ (N√N)^(-1), and, by substitution, you get x³/y³+(x³/y³)^(-1) is 52

SlawomirMlynarczyk :

Help me Andy, please. System of equations: x/y+y/x=x^y and x/y-y/x=y^x. Find x and y.

心🩷 :

after long time 😉 miss your video 🥰

DanielJBalkwill :

Since the square root of 16 can be positive or negative 4, the final value could also be -52 (cubes can be negative.)

Ron.g708 :

the sum of two cubes says a²-ab+b

ʙᴀɴ¿ ᴢᴏꜱꜱ :

w math🔥

Apo/stoli :

Pls never stop

Whiteboard Maths :

Nice Solution

Andrew DeCamp :

Missed the b squared on your sum of 2 cubes rule

DrMario67 :

no it doesn't bro✌️✌️✌️

soboi :

no

skyseaychildress :

Set u=x/y, then u>1 bc x>y>0. u²+u⁻²=14. u⁴+u⁰=14u². u⁴-14u²+1=0. Ignoring extraneous roots, u=2+sqrt(3). u³+u⁻³=(2+sqrt3)³+(2+sqrt3)⁻³=[(2³+3(2)²(sqrt3)+3(2)(sqrt3)²+(sqrt3)³]+[(2³+3(2)²(sqrt3)+3(2)(sqrt3)²+(sqrt3)³]⁻¹=[8+12sqrt3+18+3sqrt3]+[8+12sqrt3+18+3sqrt3]⁻¹=[26+15sqrt3]+[26+15sqrt3]⁻¹=[(26+15sqrt3)²+1]/[26+15sqrt3]=[26²+(2)(26)(15sqrt3)+15²(3)+1]/[26+15sqrt3]=[1352+780sqrt3]/[26+15sqrt3]=[(1352+780sqrt3)(26-15sqrt3)]/[26²-(15²)(3)]=[(1352)(26)-(1352)(15sqrt3)+(780sqrt3)(26)-(780)(15)(3)]/[1]=52. Because u³+u⁻³=52, so to does x³y⁻³+y³x⁻³=52

user1720463266748 :

Ddddxxddddxxdccffcfcm

sharkysik :

@владік

Shawan :

@

Raqif Mehdiyev425 :

👍👍👍