Find $abc$ where $a^2+b^2+c^2=144$ and $ab+bc+ca=144$ Total surface area of a cuboid is 288 sq.cm. and length of a diagonal of it is 12 cm. Find its volume.
This is the question. We know that diagonal =$a^2+b^2+c^2$ and surface area =$2(ab+bc+ca)$ and volume =$abc$ . So finally we have to find the value of $abc$ where $$a^2+b^2+c^2=144$$ and $$ ab+bc+ca =144 .$$ Somebody please help me.
 A: By the rearrangement inequality, if $a,b,c>0$ we have
$$ a^2+b^2+c^2 \geq ab+ac+bc $$
and equality occurs only at $a=b=c$. So, long story short, your constraint ensure that the cuboid is actually a cube (it has the shortest diagonal length for a given surface area).
A: Note that 
$$(a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2)-2(ab+bc+ac)=2(144-144)=0$$
which implies that $a=b=c$.
A: Hint:
$$a^2+b^2+c^2=ab+bc+cc \Leftrightarrow a=b=c \Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0$$
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$$
\left.\begin{array}{rcrcrcl}
\ds{a^{2}} & \ds{+} & \ds{b^{2}} & \ds{+} & \ds{c^{2}} & \ds{=} & \ds{144}
\\
\ds{ab} & \ds{+} & \ds{bc} & \ds{+} & \ds{ca} & \ds{=} & \ds{144}
\end{array}\right\}\,,\qquad
\mbox{Lets}\quad
\left\{\begin{array}{rcl}
\ds{\mathbf{u}} & \ds{\equiv} &
\pars{\begin{array}{c}\ds{a} \\ \ds{b}\\ \ds{c}\end{array}}
\\[2mm]
\mathsf{M} & \ds{\equiv} &
\ds{\pars{%
\begin{array}{ccc}
\ds{0} & \ds{1} & \ds{0}
\\
\ds{0} & \ds{0} & \ds{1}
\\
\ds{1} & \ds{0} & \ds{0}
\end{array}}}
\end{array}\right.
$$

The above $\ds{a,b,c}$ conditions are equivalent to:
$$
\mathbf{u}^{T}\mathbf{u} = \pars{144}\quad\mbox{and}\quad
\mathbf{u}^{T}\mathsf{M}\mathbf{u} = \pars{144}\qquad
\stackrel{\mbox{Lagrange Mult.}}{\implies}
\qquad
\mathsf{M}\mathbf{u} = \lambda\mathbf{u}
$$
The only $\ds{\mathbf{u}}$ eigenvector with real components is

$\ds{\mathbf{u} \propto
\pars{\begin{array}{ccc}\ds{1} \\ \ds{1} \\ \ds{1} \end{array}}\implies
a = b = c\implies\verts{a} = \verts{b} = \verts{c} = 4\root{3}}$. Since
$\ds{a, b, c > 0}$:
$$
\bbox[10px,#efe,border:0.1em groove navy]{a = b = c = 4\root{3}}
$$
\begin{align}
\end{align}
