anyone know how?

[vagrant briar]

https://tenor.com/view/anime-kanna-kamui-sit-sad-gif-18391892416812948817

[prime oracle]

omg she is so cute 😈

anime name?

[vagrant briar]

Miss Kobayashi's Dragon Maid

[prime oracle]

nice

let a =

let b =

slight correction

[vagrant briar]

but it won't work if it has a 9 at the front

[prime oracle]

then let nine + cube root 3 be a and 3 cube root 3 as b

just more long winded

wait

what question is this from

[vagrant briar]

what do you mean

[prime oracle]

like what is this paper set from

do you have a link to it

<@&286206848099549185>

[vagrant briar]

not cause it is a PDF on my pc

[prime oracle]

can i see the name what it is called

btw

this is the faster way of doing it

then times it by cube root 9 to the power of 2

to only get one cube root

rationalise again

no that wont work

honestly i have no idea

[vagrant briar]

that is the book, the question is 11.4.3

[vagrant briar]

same

[vagrant briar]

Don't worry, I'll win

[prime oracle]

yes

keep me updated

im interested

[vagrant briar]

okay, I will

[ripe bane]

@vagrant briar @prime oracle

Let $a=3, b=\sqrt[3]{3}$

[warm cloak]

Civil Service Pigeon

[ripe bane]

Then, the denominator is just $$a^2-ab+b^2$$

[warm cloak]

Civil Service Pigeon

[ripe bane]

Now, consider the formula ||a^3 + b^3 = (a+b)(a^2-ab+b^2)||

[vagrant briar]

https://tenor.com/view/sousou-no-frieren-frieren-beyond-journey-frieren-anime-frieren-book-drop-book-dropping-gif-5586080115796339217

ok, I got it

@ripe bane @prime oracle thank both of you for helping

[prime oracle]

men

[vagrant briar]

I start to feel I am a fool

[willow cobalt]

Try rationalizing $\frac{11}{9-3\sqrt[3]{3}+2\sqrt[3]{9}}$. That will make you suffer.

[warm cloak]

SWR

[prime oracle]

how do you know this is doable

have you had much practice

[willow cobalt]

Because I am clever

[prime oracle]

wdym

must be intuition from doing so many questions

[willow cobalt]

I know it is doable because I am a clever person

[prime oracle]

also civil is wrong

surely

[willow cobalt]

civil?

[prime oracle]

ye i dont get it

[willow cobalt]

oh

pigeon

No he got it 100% right

[prime oracle]

square of cube root of 3 ?

[willow cobalt]

(numerator should have been 6, not 11. Not that it changes much)

[willow cobalt]

$\left(\sqrt[3]{3}\right)^2$

[warm cloak]

SWR

[prime oracle]

??????

[willow cobalt]

Not $\sqrt{\sqrt[3]{3}}$

[warm cloak]

SWR

[willow cobalt]

Yeah. That's true. Denominator is $9-3\sqrt[3]{3}+\sqrt[3]{9}$

[warm cloak]

SWR

[prime oracle]

ok

so

how is that equal to that denominator

@willow cobalt

sorry its like 1 am

imma go sleep

[willow cobalt]

Because this is the denominator

[prime oracle]

so its not equal

oh right i got confused at what you were saying

so its still unsolved ig

[willow cobalt]

The point is, now that you have a and b, you can now use the sum of cubes relation

[prime oracle]

oh i see

also this do you personally know the solution

you get a really nasty surd

[willow cobalt]

yes. I do.

It takes some work, and requires solving a system of equations, but it's doable

Any algebraic number can be rationalized

[prime oracle]

ok smart man