Algebra

From the first equation we have that

$(b + c)^{2} = (56 - a)^{2} \Longrightarrow b^{2} + c^{2} + 2bc = 3136 - 112a + a^{2}$ .

Substituting $b^{2} + c^{2} = 1344 - a^{2}$ and $a^{2} = bc$ from the 2nd and 3rd equations yields

$1344 - a^{2} + 2a^{2} = 3136 - 112a + a^{2} \Longrightarrow 112a = 3136 - 1344 = 1792 \Longrightarrow$

$a = 16 \Longrightarrow abc = a \times a^{2} = 16^{3} = \boxed{4096}$ .

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Note: With $a = 16$ , we have that $b + c = 40, b^{2} + c^{2} = 1088$ and $bc = 256$ .

So $(b - c)^{2} = b^{2} + c^{2} - 2bc = 1088 - 512 = 576 \Longrightarrow b - c = \pm 24$ .

If $b - c = 24$ then $2b = (b + c) + (b - c) = 40 + 24 = 64 \Longrightarrow b = 32, c = 8$ .

If $b - c = -24$ then $2b = (b + c) + (b - c) = 40 - 24 = 16 \Longrightarrow b = 8, c = 32$ .

$a+b+c=56$

=> $(a+b+c)^{2}$ = $a^{2}$ + $b^{2}$ + $c^{2}$ + $2ab$ + $2bc$ + $2ca$

=> $(56)^{2}$ = $a^{2}$ + $b^{2}$ + $c^{2}$ + $2ab$ + $2bc$ + $2ca$ ................. $(1)$

also, $a^{2}$ + $b^{2}$ + $c^{2}$ = 1344 ................... $(2)$

putting value of $(2)$ in $(1)$ we get,

=> $3136$ = $1344$ + $2(ab+bc+ca)$

=> $1792$ = $2(ab+bc+ca)$

=> $896$ = $(ab+bc+ca)$

since, $a^{2}$ = $bc$

=> $896$ = $ab+a^{2}+ca$

=> $896$ = $a(a+b+c)$

=> ( $\frac{896}{56}$ ) = $a$ ............................................. since, $a+b+c = 56$

=> $a$ = $16$

=> $abc$ = $a.a^{2}$ .......................................... since, $a^{2} = bc$

=> $abc$ = $a^{3}$ = $16^{3}$ = $4096$

We know that , (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca) 》56^{2}=1344+2× (ab+a^{2}+ca) 》3136-1344=2a (a+b+c) 》1792=2a×56 》112a=1792 》a=16 then, abc =a×a^{2} =a^{3 } =16^{3} =4096

Consider

$\begin{aligned} (a+b+c)^2 & = a^2+b^2+c^2 + 2(ab+bc+ca) \\ 56^2 & = 1344 + 2(ab+bc+ca) \\ ab + bc + ca & = \frac {56^2-1344}2 = 896 \end{aligned}$

From $a^2 = bc$ , we note that $b$ , $a$ and $c$ are in a geometric progression . Let $b = \dfrac ar$ and $c=ar$ , where $r$ is the common ratio of the geometric progression. Then we have:

$\begin{cases} a + b + c = 56 & \implies a + \dfrac ar + ar = 56 & \implies 1 + \dfrac 1r + r = \dfrac {56}a & ...(1) \\ ab + bc + ca = 896 & \implies \dfrac {a^2}r + a^2 + a^2r = 896 & \implies 1 + \dfrac 1r + r = \dfrac {896}{a^2} & ...(2) \end{cases}$

From $(1)=(2): \ \dfrac {56}a = \dfrac {896}{a^2} \implies a = \dfrac {896}{56} = 16$ . Now, $abc = a^3 = 16^3 = \boxed{4096}$ .