# 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
5 years, 5 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?

- 5 years, 5 months ago

yup, it's right

- 5 years, 5 months ago

Do like and reshare this for others (not for my sake). Thanks!

- 5 years, 5 months ago

Note @Mehul Arora @Vaibhav Prasad , $a,b$ are real numbers and not just integers. Cheers!

- 5 years, 5 months ago

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 $(a^{2} + b^{2} + 2ab +1 +a+b$)

- 5 years, 5 months ago

Sorry , it was a typing mistake. Thanks for correcting me.

- 5 years, 5 months ago

Please explain the second last step

- 5 years, 5 months ago

$(a+b+\sqrt{2ab})(a+b-\sqrt{2ab})=(a-1)(b-1) \\ \Rightarrow (a+(b+\sqrt{2ab}))(b+(a-\sqrt{2ab}))=(a-1)(b-1)$

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

$\Rightarrow b+\sqrt{2ab} = -1 \quad a-\sqrt{2ab}=-1$

- 5 years, 5 months ago

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

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 $a =-1$.

- 5 years, 5 months ago

Nice , that crucial substitution really helped!

- 5 years, 5 months ago

That identity is very useful it is just a result derived from another identity.

- 5 years, 4 months ago

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. - 5 years, 5 months ago

Sir , note that $a+b=-2$ is also a solution. See my and Krishna's solution. Thanks!

- 5 years, 5 months ago

There is one special case when $a+b=-2$.

- 5 years, 5 months ago

- 5 years, 5 months ago

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

- 5 years, 5 months ago

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

- 5 years, 5 months ago

- 5 years, 5 months ago

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

- 5 years, 5 months ago

How about $a=-1$ and $b=-1$

- 5 years, 5 months ago

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

- 5 years, 5 months ago

See my bash , no need to compute $a,b$ :)

- 5 years, 5 months ago

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.

- 5 years, 5 months ago

That's right.

- 5 years, 5 months ago

I know.

- 5 years, 5 months ago

But that's incomplete. -_-

- 5 years, 5 months ago

Somewhat. But I laid the basis, didn't I ? :P

- 5 years, 5 months ago

Yeah you have a nice different idea. I am trying a different approach... xD

- 5 years, 5 months ago

Okay! So Cheers! xD

- 5 years, 5 months ago

See my bash , no need to compute $a,b$ :)

- 5 years, 5 months ago

a+b can be 1 also

- 5 years, 5 months ago

Yeah it can be $1$. So?

- 5 years, 5 months ago

LOL xD

- 5 years, 5 months ago