1 equation , 2 variables.Solvable?
If , Find value of
Nice solutions are always welcome!
Note by
Nihar Mahajan
6 years, 1 month ago
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So , here is my trial:
a3+b3+3ab−1=0(a+b)3−3ab(a+b)+3ab−1=0(a+b)3−3ab(a+b−1)−1=0(a+b)3−13=3ab(a+b−1)(a+b−1)(a2+b2+2ab+a+b+1)=3ab(a+b−1)
Let a+b−1=0
⇒a2+b2+2ab+a+b+1=3aba2+b2−ab+a+b+1=0a2+b2−(a−1)(b−1)=0a2+b2=(a−1)(b−1)(a+b)2−2ab=(a−1)(b−1)(a+b+2ab)(a+b−2ab)=(a−1)(b−1)⇒b+2ab=−1a−2ab=−1⇒a+b=−2a+b=1
Is it right?
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yup, it's right
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Do like and reshare this for others (not for my sake). Thanks!
Note @Mehul Arora @Vaibhav Prasad , a,b are real numbers and not just integers. Cheers!
Hey ,in the 5th step ,from where has -1 arrived in the second bracket I mean when you used the formula a3−b3 the correct equation should have been this (a2+b2+2ab+1+a+b)
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Sorry , it was a typing mistake. Thanks for correcting me.
Please explain the second last step
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(a+b+2ab)(a+b−2ab)=(a−1)(b−1)⇒(a+(b+2ab))(b+(a−2ab))=(a−1)(b−1)
If you compare the factors of L.H.S and R.H.S , correspondingly you will get:
⇒b+2ab=−1a−2ab=−1
a3+b3+c3−3abc=21(a+b+c)((a−b)2+(b−c)2+(c−a)2)
Now substitute c=−1 and then L.H.S equals 0
0=(a+b−1)((a−b)2+(a+1)2+(b+1)2)
From here
a+b=1 from the first and
a+b=−2 from second (easily solvable)
EDIT :- the second equation has only one solution when a=−1.
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Nice , that crucial substitution really helped!
That identity is very useful it is just a result derived from another identity.
The answer is a+b=1.
I plotted f(a)=a3+b3+3ab−1 for b=−10,−7.5,0,7.5,10, thinking that there will be infinite solutions. But I discovered that the roots to f(a) are respectively (a=11,8.5,1,-6.5,-9) making a+b=1. See the graph here.
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Sir , note that a+b=−2 is also a solution. See my and Krishna's solution. Thanks!
There is one special case when a+b=−2.
@Calvin Lin @siddharth bhatt @Pi Han Goh @Brian Charlesworth @Curtis Clement @Azhaghu Roopesh M @Mehul Arora @Chew-Seong Cheong @Prasun Biswas @everyone no need of help now.
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What do you need help with? You solved the problem didn't you? :)
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Actually , I got it just after I posted this note. Since you have nice methods and approaches , I mentioned you.
1 :- The answer
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I am not asking for the answer. I am asking for an elegant accurate solution with proof. -_-
How about a=−1 and b=−1
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I am not asking for the answer. I am asking for an elegant accurate solution with proof. -_-
See my bash , no need to compute a,b :)
Let a+b=x Now If, x=1, Then the Equation becomes a+b ^ 3 Thena+b=1 :D
P.S. I'm just too lazy. Idk What I am typing right now.
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That's right.
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I know.
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But that's incomplete. -_-
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Somewhat. But I laid the basis, didn't I ? :P
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Yeah you have a nice different idea. I am trying a different approach... xD
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Okay! So Cheers! xD
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See my bash , no need to compute a,b :)
a+b can be 1 also
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Yeah it can be 1. So?
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LOL xD