1 equation , 2 variables.Solvable?

a3+b3+3ab=1\large a^3+b^3+3ab=1

If a,bRa,b \in R , Find value of a+ba+b

Nice solutions are always welcome!

Note by Nihar Mahajan
4 years, 7 months ago

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So , here is my trial:

a3+b3+3ab1=0(a+b)33ab(a+b)+3ab1=0(a+b)33ab(a+b1)1=0(a+b)313=3ab(a+b1)(a+b1)(a2+b2+2ab+a+b+1)=3ab(a+b1)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+b10a+b-1 \neq 0

a2+b2+2ab+a+b+1=3aba2+b2ab+a+b+1=0a2+b2(a1)(b1)=0a2+b2=(a1)(b1)(a+b)22ab=(a1)(b1)(a+b+2ab)(a+b2ab)=(a1)(b1)b+2ab=1a2ab=1a+b=2a+b=1\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 - 4 years, 7 months ago

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yup, it's right

Vaibhav Prasad - 4 years, 7 months ago

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

Nihar Mahajan - 4 years, 7 months ago

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

Nihar Mahajan - 4 years, 7 months ago

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Hey ,in the 5th step ,from where has -1 arrived in the second bracket I mean when you used the formula a3b3a^{3} - b^{3} the correct equation should have been this (a2+b2+2ab+1+a+b(a^{2} + b^{2} + 2ab +1 +a+b)

Aman Real - 4 years, 7 months ago

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

Nihar Mahajan - 4 years, 7 months ago

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Please explain the second last step

Aman Real - 4 years, 7 months ago

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(a+b+2ab)(a+b2ab)=(a1)(b1)(a+(b+2ab))(b+(a2ab))=(a1)(b1)(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.SL.H.S and R.H.SR.H.S , correspondingly you will get:

b+2ab=1a2ab=1\Rightarrow b+\sqrt{2ab} = -1 \quad a-\sqrt{2ab}=-1

Nihar Mahajan - 4 years, 7 months ago

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a3+b3+c33abc=12(a+b+c)((ab)2+(bc)2+(ca)2)\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=1c = -1 and then L.H.S equals 0

0=(a+b1)((ab)2+(a+1)2+(b+1)2) \displaystyle 0 = (a+b-1)( (a-b)^2 + (a+1)^2 + (b+1)^2)

From here

a+b=1a+b = 1 from the first and

a+b=2a+b = -2 from second (easily solvable)

EDIT :- the second equation has only one solution when a=1a =-1.

Krishna Sharma - 4 years, 7 months ago

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Nice , that crucial substitution really helped!

Nihar Mahajan - 4 years, 7 months ago

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

shivamani patil - 4 years, 6 months ago

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

I plotted f(a)=a3+b3+3ab1f(a) = a^3+b^3+3ab-1 for b=10,7.5,0,7.5,10b = -10, -7.5, 0, 7.5, 10, thinking that there will be infinite solutions. But I discovered that the roots to f(a)f(a) are respectively (a=11,8.5,1,-6.5,-9) making a+b=1\boxed{a+b = 1}. See the graph here.

Chew-Seong Cheong - 4 years, 7 months ago

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

Nihar Mahajan - 4 years, 7 months ago

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

Krishna Sharma - 4 years, 7 months ago

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

Curtis Clement - 4 years, 7 months ago

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

Nihar Mahajan - 4 years, 7 months ago

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1 :- The answer

Mehul Arora - 4 years, 7 months ago

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

Nihar Mahajan - 4 years, 7 months ago

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

Vaibhav Prasad - 4 years, 7 months ago

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

Nihar Mahajan - 4 years, 7 months ago

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

Nihar Mahajan - 4 years, 7 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 - 4 years, 7 months ago

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That's right.

Nihar Mahajan - 4 years, 7 months ago

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I know.

Mehul Arora - 4 years, 7 months ago

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@Mehul Arora But that's incomplete. -_-

Nihar Mahajan - 4 years, 7 months ago

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

Mehul Arora - 4 years, 7 months ago

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

Nihar Mahajan - 4 years, 7 months ago

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@Nihar Mahajan Okay! So Cheers! xD

Mehul Arora - 4 years, 7 months ago

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@Mehul Arora See my bash , no need to compute a,ba,b :)

Nihar Mahajan - 4 years, 7 months ago

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a+b can be 1 also

Pranjit Handique - 4 years, 7 months ago

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Yeah it can be 11. So?

Nihar Mahajan - 4 years, 7 months ago

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LOL xD

Mehul Arora - 4 years, 7 months ago

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