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

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

Nice solutions are always welcome!

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

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

Nice solutions are always welcome!

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TopNewestSo , 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, 6 months ago

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– Aman Real · 1 year, 6 months ago

Please explain the second last stepLog in to reply

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, 6 months ago

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-1arrived 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, 6 months agoLog in to reply

– Nihar Mahajan · 1 year, 6 months ago

Sorry , it was a typing mistake. Thanks for correcting me.Log in to reply

@Mehul Arora @Vaibhav Prasad , \(a,b\) are real numbers and not just integers. Cheers! – Nihar Mahajan · 1 year, 6 months ago

NoteLog in to reply

– Vaibhav Prasad · 1 year, 6 months ago

yup, it's rightLog in to reply

– Nihar Mahajan · 1 year, 6 months ago

Do like and reshare this for others (not for my sake). Thanks!Log in to reply

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

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, 6 months ago

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– Shivamani Patil · 1 year, 5 months ago

That identity is very useful it is just a result derived from another identity.Log in to reply

– Nihar Mahajan · 1 year, 6 months ago

Nice , that crucial substitution really helped!Log in to reply

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, 6 months agoLog in to reply

– Krishna Sharma · 1 year, 6 months ago

There is one special case when \(a+b=-2\).Log in to reply

– Nihar Mahajan · 1 year, 6 months ago

Sir , note that \(a+b=-2\) is also a solution. See my and Krishna's solution. Thanks!Log in to reply

a+b can be 1 also – Pranjit Handique · 1 year, 6 months ago

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– Nihar Mahajan · 1 year, 6 months ago

Yeah it can be \(1\). So?Log in to reply

– Mehul Arora · 1 year, 6 months ago

LOL xDLog in to reply

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, 6 months ago

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– Nihar Mahajan · 1 year, 6 months ago

That's right.Log in to reply

– Mehul Arora · 1 year, 6 months ago

I know.Log in to reply

– Nihar Mahajan · 1 year, 6 months ago

But that's incomplete. -_-Log in to reply

– Mehul Arora · 1 year, 6 months ago

Somewhat. But I laid the basis, didn't I ? :PLog in to reply

– Nihar Mahajan · 1 year, 6 months ago

Yeah you have a nice different idea. I am trying a different approach... xDLog in to reply

– Mehul Arora · 1 year, 6 months ago

Okay! So Cheers! xDLog in to reply

– Nihar Mahajan · 1 year, 6 months ago

See my bash , no need to compute \(a,b\) :)Log in to reply

How about \(a=-1\) and \(b=-1\) – Vaibhav Prasad · 1 year, 6 months ago

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– Nihar Mahajan · 1 year, 6 months ago

See my bash , no need to compute \(a,b\) :)Log in to reply

– Nihar Mahajan · 1 year, 6 months ago

I am not asking for the answer. I am asking for an elegant accurate solution with proof. -_-Log in to reply

1 :- The answer – Mehul Arora · 1 year, 6 months ago

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– Nihar Mahajan · 1 year, 6 months ago

I am not asking for the answer. I am asking for an elegant accurate solution with proof. -_-Log in to reply

@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. – Nihar Mahajan · 1 year, 6 months ago

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– Curtis Clement · 1 year, 6 months ago

What do you need help with? You solved the problem didn't you? :)Log in to reply

– Nihar Mahajan · 1 year, 6 months ago

Actually , I got it just after I posted this note. Since you have nice methods and approaches , I mentioned you.Log in to reply