Working With A Ratio

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[@echandler]

once you have (a+b)/(a-b) = 5 = 5/1, you can use componendo-dividendo again or *reverse* componendo-dividendo.
(5+1)/2 = 3, (5-1)/2 = 2 thus (a+b)/(a-b) = (3+2)/(3-2) hence a/b = 3/2.

[@williamperez-hernandez3968]

Typo at 7:40. The coefficient of u^2 is -189.

[@MrGeorge1896]

62 b³ ((a/b)³ + 3 (a/b)) = 63 a³ ((b/a)³ + 3 (b/a)) and now with c = a/b we get:
62c³ - 189c² + 186x -63 = 0 (there is a typo in the video: -129 instead of -189)
(2c - 3) (31c² - 48c + 21) = 0 with one real solution c= 3/2 and two complex ones.

[@msmbpc24]

3/2

[@phill3986]

😊😊😊👍👍👍

[@jackkalver4644]

If you let a/b=u, then the equation becomes (u^3+3u)/(3u^2+1)=63/62, or u=triscoth(63/62). The only problem is hunting a rational root. Just solve 62(u^3+3u)=63(3u^2+1)

[@nasrullahhusnan2289]

By inspection (a,b)=(3,2)
Thus a/b=3/2
Check:
a³+3ab²=3³+3(3)2²
=27+36=63
b³+3a²b=2³+3(3²)2
=8+54=62

[@DonEnsley-mathdrum]

problem
(a³+3ab²)/(b³+3a²b) = 63/62
This is a homogeneous system of degree 3.
Let
b = ka
, where k is a real number.
Replacing gives us
(1+3k²)/[k(k² +3)] = 63/62
63k³-186k² + 189 k-62 =0
(3k-2)(21k-48k+31)=0
3k-2 = 0
k = ⅔
b = ⅔ a
All ordered pairs (a,b) that solve this system are
(a,b) ∈ { (a, ⅔ a), a ∈ ℝ }
which means that a/b is
3/2.
answer
3/2
(a,b) ∈ { (a, ⅔ a), a ∈ ℝ }

[@maxhagenauer24]

The x/y = z/w => (x+y)/(x-y) = (z+w)/(z-w) works because
x/y = z/w
x = zy/w
x+y = y+(zy)/w
(x+y)/(x-y) = ((zy/w)+y)/(x-y)
Knowing that y = xw/z, we can replace
(x+y)/(x-y) = xw(z+w)/(zx(1-w/z))
(x+y)/(x-y) = (z+w)/(z-w)