I wonder if there are some complex solutions too. I think what would change is that at the a^3 = -b^3, instead of just a=-b we also get potential solutions of a = (1+i√3)/2 b and a = (1-i√3)/2 b. (This corresponds to e^iπ, e^iπ/3, and e^5iπ/3 = e^-iπ/3; and is related to the third roots of unity). From here I'm not exactly sure where to go though. But I suspect the answers would have the same magnitude.
@russelltownsend6105
2 жыл бұрын
Love your logic and clarity thanks heaps!
@nexusmas9
2 жыл бұрын
Bro this is insane , I don't get how you could solve letters . Mashallah ❤
@sanjaysurya6840
2 жыл бұрын
🤣😊
@crustyoldfart
2 жыл бұрын
At the risk of appearing like an untutored peasant, my reaction to this kind of puzzle is that it can barely be considered any more than a storm in teacup. Clearly, a,b,c,d are homogeneous and any solution for one of them can be equally asserted for each of the others. Accordingly we would expect there to be four interchangeable solution sets. FOUR ? Yes, if 0 is considered non-trivial. I don't have the inclination to rummage round finding specific complex solutions, but it is reasonable to suppose there might be at least one such, as others have pointed out. For instance solving of any one of the four individual equations there are three solutions for each - of the form A, A/2 +/- sqrt(3)/2*i. For example, solving for d : A= ( b - a)^(1/3) -> d = { A, A/2*( -1+sqrt(3)/2*i), A/2*(-1-sqrt(3)/2*i) }
@kanishkaasde
2 жыл бұрын
Supercalifragilisticexpialidocious
@necro5379
2 жыл бұрын
Great🔥
@xactxx
2 жыл бұрын
In 6:50, you had earlier proved that both pairs (a,c) and (b,d) have the same sign. That means that ac and bd must be positive. So both a^2+ac+b^2+1 and b^2+bd+d^2+1 must be both positive immediately.
@shankhadeepghosh8574
2 жыл бұрын
Unique thought process👍❤️
@bait6652
2 жыл бұрын
Is it sufficient o note the longer term is pos...so The sign(linear terms) equal...thus sgn(d-b)=sgn(b-d) which should = 0 ..thus b=d , a=c, sub in sum a =-b
@alboris8203
2 жыл бұрын
Cool
@តន្ដ្រីស្រុកខ្ញុំ
2 жыл бұрын
I like hard questions
@زينالعابدينماجدمحمد
2 жыл бұрын
THANK YOU so much ❤
@prometheus3899
2 жыл бұрын
a³+b³+c³+d³ is 0 and subtraction 2 from 1 a³-b³=-b-d c³-d³=-d-b from eqn 3 and 4 Thus a³-b³ = c³-d³ ; a³+d³=b³+c³ =0 so a=-d b=-c a³=-2b -b³= 2a Solve this to get reqd answers
@isaiaherb6116
2 жыл бұрын
I don’t think the subtraction step works. The c’s and a’s are constructive.
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