This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .

Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.

Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold

- bulleted - list

bulleted

list

1. numbered 2. list

numbered

list

Note: you must add a full line of space before and after lists for them to show up correctly

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.

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

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewesta+b can be 1 also

Log in to reply

Yeah it can be $1$. So?

Log in to reply

LOL xD

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.

Log in to reply

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

Log in to reply

Sir , note that $a+b=-2$ is also a solution. See my and Krishna's solution. 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 $a =-1$.

Log in to reply

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

Log in to reply

Nice , that crucial substitution really helped!

Log in to reply

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?

Log in to reply

Please explain the second last step

Log in to reply

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

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

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

Log in to reply

Hey ,in the 5th step ,from where has

-1arrived in the second bracket I mean when you used the formula $a^{3} - b^{3}$ the correct equation should have been this$(a^{2} + b^{2} + 2ab +1 +a+b$)Log in to reply

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

Log in to reply

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

Log in to reply

yup, it's right

Log in to reply

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

Log 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.

Log in to reply

That's right.

Log in to reply

I know.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

$a,b$ :)

See my bash , no need to computeLog in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

1 :- The answer

Log in to reply

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.

Log in to reply

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

Log in to reply

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

Log in to reply