If $a^3 + b^3 +3ab = 1$, find $a+b$

Question

Given that the real numbers $a,b$ satisfy $a^3 + b^3 +3ab = 1$, find $a+b$.

I tried to factorize it but unable to do it.

Answer 1

There are a continuum of solutions to $$ a^3+b^3+3ab=1 $$ Suppose that $$ x=a+b $$ then $$ \begin{align} 1 &=a^3+(x-a)^3+3a(x-a)\\ &=x^3-3ax^2+3a^2x+3ax-3a^2 \end{align} $$ which means $$ \begin{align} 0 &=(x-1)\left(x^2+(1-3a)x+3a^2+1\right) \end{align} $$ So either $x=1$ irregardless of $a$, or $$ x=\frac{3a-1\pm(a+1)\sqrt{-3}}{2} $$ Thus, other than $x=1$, the only real $x$ is $-2$, which comes from $a=-1$.

That is, the only two real values of $a+b$ are $1$ and $-2$.

Answer 2

HINT:

$$(a+b)^3=a^3+3a^2b+3ab^2+b^3=a^3+3ab(a+b)+b^3$$

Or see this question, especially my answer to it.

Answer 3

Hint:

$$a^3+b^3+3ab=(a+b)^3-3ab(a+b)+3ab\ldots\ldots$$

Answer 4

Hint: \begin{align} x^3+y^3+z^3-3xyz& =(x+y+z)(x^2+y^2+z^2-xy-xz-yz) \\ &=(x+y+z)\left(\frac{(x-y)^2+(x-z)^2+(y-z)^2}{2}\right) \end{align}

Solution: $$0=a^3+b^3+(-1)^3-3(a)(b)(-1)=(a+b-1)\left(\frac{(a-b)^2+(a+1)^2+(b+1)^2}{2}\right)$$ so $a+b=1$ or $a=b=-1$. The latter gives $a+b=-2$.

Answer 5

The three solutions of the equation $a^3+b^3+3ab=1$ are

$$b_1=1-a,$$ $$b_2,b_3 = \frac{1}{2}\left(a-1\pm i\sqrt{3}(1+a)\right).$$

If $a\neq -1$, since $a,b\in\mathbb{R}$, we must have $a+b=a+b_1=1$. If $a=-1$, then the imaginary part of $b_2,b_3$ vanishes and we find two solutions for $(a,b)$: $(-1,-1)$ and $(-1,2)$ so $a+b$ is -2 or 1 respectively.

Answer 6

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$

$$ \pars{a + b}^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3} = 1 - 3ab + 3a^{2}b + 3ab^{2} = 1 + 3ab\pars{-1 + a + b} $$ $$ x^{3} -3abx + 3ab - 1 = 0\quad\mbox{where}\quad x \equiv a + b $$ $$ \pars{x - 1}\pars{x^{2} + x + 1} - 3ab\pars{x - 1} = 0 $$ One solution is $\color{#ff0000}{\large x = a + b = 1}$.

When $x \not=1$ $\pars{~a + b \not= 1~}$: $$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! 0 = x^{2} + x + \pars{1 - 3ab} = x^{2} + x + 1 - 3a\pars{x - a} = x^{2} + \pars{1 - 3a}x + \pars{1 + 3a^{2}} = 0 \tag{1} $$ This equation discriminant $\Delta$ is given by: $$ \Delta = \pars{1 - 3a}^{2} - 4\pars{1 + 3a^{2}} = -3a^{2} - 6a - 3 = -3\pars{a + 1}^{2} \leq 0 $$

Since $x = \pars{a + b} \in {\mathbb R}$, this analysis doesn't yield any other solution when $a \not= -1$. When $a = -1$, Eq. $\pars{1}$ has the double root $x_{\pm} = \pars{3a - 1}/2 = -2$. Then,

$$ \mbox{the solutions are}\quad {\large\left\lbrace% \begin{array}{rcl} a + b & = & \phantom{-}1 \\[1mm] \mbox{and}\quad a + b & = & -2 \quad\mbox{when}\quad a = -1 \end{array}\right.} $$

Answer 7

As stated, there is no unique answer.

Let x = a + b, c = 3ab. Then

a³ + b³ + 3ab(a + b) = (a + b)³, i.e.

1 = a³ + b³ + 3ab = (a + b)³ - 3ab(a + b - 1), i.e.

x³ - cx + c - 1 = 0, i.e.

(x - 1)(x² + x - c + 1) = 0, i.e.

x = 1 or x = ½ [± √(4c - 3) - 1], if c ≥ ¾.