a,b and c real numbers

Distributing gives:

$ab + ac = 152$
$ab + bc = 162$
$ac + bc = 170$

Subtracting the first from the second gives:

$bc - ac = 10$
$ac + bc = 170$

Solving for $bc$ , and subsequently $ab$ and $ac$ :
$bc = 90,\, ab = 72,\,ac=80$

Then, $abc = \sqrt{a^{2}b^{2}c^{2}} = \sqrt{ab*ac*bc} = \sqrt{90 * 72 * 80} = 720$

$\begin{cases} a(b+c) = ab+ca = 152 &...(1) \\ b(a+c) = ab+bc = 162 & ...(2) \\ c(a+b) = ca+bc = 170 &...(3) \end{cases}$

$\begin{aligned} (1)+(2)-(3): \quad 2ab & = 144 \\ ab & = 72 & ...(4) \end{aligned}$

$\begin{aligned} (1): \quad ca & = 152-72 = 80 & ...(5) \\ (2): \quad bc & = 162-72 = 90 & ...(6) \end{aligned}$

$\begin{aligned} (4)(5)(6): \quad (abc)^2 & = 72\cdot 80 \cdot 90 \\ abc & = \boxed{720} \end{aligned}$

ab+ac=152. A ba+bc=162. B ca+cb=170. C

Therefore, a=152/b+c, b=162/a+c, c=170/a+b

Taking A and substituting b: a(162/a+c) + ac = 152, 162a/a+c + ac = 152, 162a + ac(a+c) = 152a + 152c, 162a + ca^2 + ac^2 = 152a + 152c, ac^2 + ca^2 + 10a - 152c = 0

Taking C and substituting b: ca + c(162/a+c) = 170, ca + 162c/a+c = 170, ca(a+c) + 162c = 170a + 170c, ca^2 + ac^2 + 162c = 170a + 170c, ac^2 + ca^2 - 8c - 170 = 0

Clearly: 10a - 152c = -8c - 170a, 180a = 144c, 15a = 12c, 5a = 4c, a = 4c/5

Taking b=162/a+c: b = 162/(4c/5 + c), b = 162/(9c/5), 9c/5 = 162/b, 9cb = 810, bc = 90, b = 90/c

Taking A: ab + ac = 152, (4c/5)(90/c) + (4c/5)c = 152, 72 + 4c^2/5 = 760, 4c^2 = 400, c^2 = 100, c = 10. (Note: question says positive integers so c can't be -10) And so using c=10, b=9, and a=8.

abc= 720 😃😃😃