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.
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.
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)
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)
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.
Typo at 7:40. The coefficient of u^2 is -189.
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.
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
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)
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)
3/2
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