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1 equation , 2 variables.Solvable?

\[\large a^3+b^3+3ab=1\]

If \(a,b \in R\) , Find value of \(a+b\)

Nice solutions are always welcome!

Note by Nihar Mahajan
1 year, 9 months ago

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So , here is my trial:

\[a^3+b^3+3ab-1=0 \\ (a+b)^3-3ab(a+b)+3ab-1=0 \\ (a+b)^3 - 3ab(a+b-1) -1=0 \\ (a+b)^3 - 1^3 = 3ab(a+b-1) \\ (a+b-1)(a^2+b^2+2ab+a+b+1)= 3ab(a+b-1)\]

Let \(a+b-1 \neq 0\)

\[\Rightarrow a^2+b^2+2ab+a+b+1= 3ab \\ a^2+b^2-ab+a+b+1=0 \\ a^2+b^2-(a-1)(b-1)=0 \\ a^2+b^2=(a-1)(b-1) \\ (a+b)^2 - 2ab = (a-1)(b-1) \\ (a+b+\sqrt{2ab})(a+b-\sqrt{2ab})=(a-1)(b-1) \\ \Rightarrow b+\sqrt{2ab} = -1 \quad a-\sqrt{2ab}=-1 \\ \Rightarrow a+b=-2 \quad a+b=1\]

Is it right? Nihar Mahajan · 1 year, 9 months ago

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@Nihar Mahajan Please explain the second last step Aman Real · 1 year, 9 months ago

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@Aman Real \[(a+b+\sqrt{2ab})(a+b-\sqrt{2ab})=(a-1)(b-1) \\ \Rightarrow (a+(b+\sqrt{2ab}))(b+(a-\sqrt{2ab}))=(a-1)(b-1)\]

If you compare the factors of \(L.H.S\) and \(R.H.S\) , correspondingly you will get:

\[\Rightarrow b+\sqrt{2ab} = -1 \quad a-\sqrt{2ab}=-1\] Nihar Mahajan · 1 year, 9 months ago

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@Nihar Mahajan Hey ,in the 5th step ,from where has -1 arrived in the second bracket I mean when you used the formula \(a^{3} - b^{3}\) the correct equation should have been this \((a^{2} + b^{2} + 2ab +1 +a+b\)) Aman Real · 1 year, 9 months ago

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@Aman Real Sorry , it was a typing mistake. Thanks for correcting me. Nihar Mahajan · 1 year, 9 months ago

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@Nihar Mahajan Note @Mehul Arora @Vaibhav Prasad , \(a,b\) are real numbers and not just integers. Cheers! Nihar Mahajan · 1 year, 9 months ago

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@Nihar Mahajan yup, it's right Vaibhav Prasad · 1 year, 9 months ago

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@Vaibhav Prasad Do like and reshare this for others (not for my sake). Thanks! Nihar Mahajan · 1 year, 9 months ago

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\(\displaystyle a^3 + b^3 + c^3 -3abc = \dfrac{1}{2} (a+b+c)( (a-b)^2 + (b-c)^2 + (c-a)^2)\)

Now substitute \(c = -1\) and then L.H.S equals 0

\( \displaystyle 0 = (a+b-1)( (a-b)^2 + (a+1)^2 + (b+1)^2)\)

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 \(a =-1\). Krishna Sharma · 1 year, 9 months ago

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@Krishna Sharma That identity is very useful it is just a result derived from another identity. Shivamani Patil · 1 year, 8 months ago

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@Krishna Sharma Nice , that crucial substitution really helped! Nihar Mahajan · 1 year, 9 months ago

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The answer is \(\space \boxed{a+b=1}\).

I plotted \(f(a) = a^3+b^3+3ab-1\) 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 \(\boxed{a+b = 1}\). See the graph here.

Chew-Seong Cheong · 1 year, 9 months ago

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@Chew-Seong Cheong There is one special case when \(a+b=-2\). Krishna Sharma · 1 year, 9 months ago

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@Chew-Seong Cheong Sir , note that \(a+b=-2\) is also a solution. See my and Krishna's solution. Thanks! Nihar Mahajan · 1 year, 9 months ago

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a+b can be 1 also Pranjit Handique · 1 year, 9 months ago

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@Pranjit Handique Yeah it can be \(1\). So? Nihar Mahajan · 1 year, 9 months ago

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@Nihar Mahajan LOL xD Mehul Arora · 1 year, 9 months ago

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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. Mehul Arora · 1 year, 9 months ago

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@Mehul Arora That's right. Nihar Mahajan · 1 year, 9 months ago

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@Nihar Mahajan I know. Mehul Arora · 1 year, 9 months ago

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@Mehul Arora But that's incomplete. -_- Nihar Mahajan · 1 year, 9 months ago

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@Nihar Mahajan Somewhat. But I laid the basis, didn't I ? :P Mehul Arora · 1 year, 9 months ago

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@Mehul Arora Yeah you have a nice different idea. I am trying a different approach... xD Nihar Mahajan · 1 year, 9 months ago

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@Nihar Mahajan Okay! So Cheers! xD Mehul Arora · 1 year, 9 months ago

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@Mehul Arora See my bash , no need to compute \(a,b\) :) Nihar Mahajan · 1 year, 9 months ago

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How about \(a=-1\) and \(b=-1\) Vaibhav Prasad · 1 year, 9 months ago

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@Vaibhav Prasad See my bash , no need to compute \(a,b\) :) Nihar Mahajan · 1 year, 9 months ago

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@Vaibhav Prasad I am not asking for the answer. I am asking for an elegant accurate solution with proof. -_- Nihar Mahajan · 1 year, 9 months ago

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1 :- The answer Mehul Arora · 1 year, 9 months ago

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@Mehul Arora I am not asking for the answer. I am asking for an elegant accurate solution with proof. -_- Nihar Mahajan · 1 year, 9 months ago

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@Nihar Mahajan What do you need help with? You solved the problem didn't you? :) Curtis Clement · 1 year, 9 months ago

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@Curtis Clement Actually , I got it just after I posted this note. Since you have nice methods and approaches , I mentioned you. Nihar Mahajan · 1 year, 9 months ago

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