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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, 11 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? · 1 year, 11 months ago

Please explain the second last step · 1 year, 11 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:

$\Rightarrow b+\sqrt{2ab} = -1 \quad a-\sqrt{2ab}=-1$ · 1 year, 11 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 $$(a^{2} + b^{2} + 2ab +1 +a+b$$) · 1 year, 11 months ago

Sorry , it was a typing mistake. Thanks for correcting me. · 1 year, 11 months ago

Note @Mehul Arora @Vaibhav Prasad , $$a,b$$ are real numbers and not just integers. Cheers! · 1 year, 11 months ago

yup, it's right · 1 year, 11 months ago

Do like and reshare this for others (not for my sake). Thanks! · 1 year, 11 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 $$a =-1$$. · 1 year, 11 months ago

That identity is very useful it is just a result derived from another identity. · 1 year, 10 months ago

Nice , that crucial substitution really helped! · 1 year, 11 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.

· 1 year, 11 months ago

There is one special case when $$a+b=-2$$. · 1 year, 11 months ago

Sir , note that $$a+b=-2$$ is also a solution. See my and Krishna's solution. Thanks! · 1 year, 11 months ago

a+b can be 1 also · 1 year, 11 months ago

Yeah it can be $$1$$. So? · 1 year, 11 months ago

LOL xD · 1 year, 11 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. · 1 year, 11 months ago

That's right. · 1 year, 11 months ago

I know. · 1 year, 11 months ago

But that's incomplete. -_- · 1 year, 11 months ago

Somewhat. But I laid the basis, didn't I ? :P · 1 year, 11 months ago

Yeah you have a nice different idea. I am trying a different approach... xD · 1 year, 11 months ago

Okay! So Cheers! xD · 1 year, 11 months ago

See my bash , no need to compute $$a,b$$ :) · 1 year, 11 months ago

How about $$a=-1$$ and $$b=-1$$ · 1 year, 11 months ago

See my bash , no need to compute $$a,b$$ :) · 1 year, 11 months ago

I am not asking for the answer. I am asking for an elegant accurate solution with proof. -_- · 1 year, 11 months ago

1 :- The answer · 1 year, 11 months ago

I am not asking for the answer. I am asking for an elegant accurate solution with proof. -_- · 1 year, 11 months ago

What do you need help with? You solved the problem didn't you? :) · 1 year, 11 months ago