Triply Complex Triples

The first equation can be expressed as:

$ab+3b=-9$

$b(a+3)=-9$

$b=\frac{-9}{a+3}$

Similarly, $c=\frac{-9}{b+3}$ .

By substitution, $c=\dfrac{-9}{\dfrac{-9}{a+3}+3} = \frac{-3(a+3)}{a}$

Multiplying $a$ , $b$ and $c$ yields:

$abc = (a)(\frac{-9}{a+3})(\frac{-3(a+3)}{a})$

$abc = \fbox{27}$ .

First, we add all of the given equations to get $\qquad ab+bc+ca+3a+3b+3c=-27 \qquad (1).$

Second, we multiply the first equation by $c,$ the second by $a,$ and the third by $b.$ We get: $\begin{aligned} abc+3bc &= -9c \\ abc + 3ca &= -9a \\ abc + 3ab &= -9b \end{aligned}$ Now we add these three equations together to get $3abc+3ab+3bc+3ca = -9a-9b-9c,$ or $abc+ab+bc+ca=-3a-3b-3c.$ Moving all of the terms to the left-hand side, we obtain $abc+ac+bc+ca+3a+3b+3c = 0.$ From equation $(1),$ we have $ab+bc+ca+3a+3b+3c = -27,$ so $abc - 27 = 0,$ and therefore $abc = \boxed{27},$ the only possible value.

first we sum up ab+3b=−9 bc+3c=−9 ca+3a=−9 yields ab+bc+ac+3(a+b+c)+27=0

and now

ab+3b=−9 multiply with c => abc+3bc=-9c

bc+3c=−9 multiply with a => abc+3ac=-9a

ca+3a=−9 multiply with b => abc+3ab=-9b and sum up yields

ab+bc+ac+3(a+b+c)+abc=0

then we conclude abc=27.

Using the formulas given, $abc = a(bc) = a(-9 - 3c) = -3(ac + 3a) = -3(-9) = 27$ . Hence, assuming at least one solution $(a,b,c)$ , we have $|abc| = |27| = \boxed{27}$ .

ab + 3b = bc + 3c. Implies that: c(b-a) = 3(a-c). So, c = 3(a-c)/(b-a) Analogously, a = 3(a-b)/(b-c) and b = 3(b-c)/(c-a).

So, |abc| = 27

since the equations are similar we can take a=b=c.(we can check it by replacing a=b, b=c, in first equation we will get the 2nd equation). Another approach is to maximize |abc| we have to take a=b=c. so we get a^2+3a=-9. =>a^2+3a+9=0 we get |a|= 3. and we know |abc|=|a||b||c| so the ans is 3 3 3=27.

(ab+3b)(bc+3c)(ca+3a)=-729

abc(a+3)(b+3)(c+3)=-729

abc(abc+3(ab+ac+bc+3a+3b+3c)+27)

abc(abc+3(-27)+27)

(abc)²-54abc+729=0 abc=27

Adding all the equations will give us:

$ab + bc + ac + 3a + 3b + 3c = -27$

Solving for $a, b \text { and } c$ from the three equations, we get:

$b(a + 3) = -9 \Rightarrow b = \frac {-9}{a + 3}$

$c(b + 3) = -9 \Rightarrow c = \frac {-9}{b + 3}$

$a(c + 3) = -9 \Rightarrow a = \frac {-9}{c + 3}$

Then, we get $abc$ from the product of the three new equations:

$abc = \frac {-729}{(a + 3)(b + 3)(c + 3)}$

$abc = \frac {-729}{abc + 3ab + 3bc + 3ac + 9a + 9b + 9c +27}$

$abc = \frac {-729}{abc + 27 +3(ab + bc + ac + 3a + 3b + 3c)}$ ,

but $ab + bc + ac + 3a + 3b + 3c = -27$ . Therefore:

$abc = \frac {-729}{abc + 27 + 3(-27)}$

$abc = \frac {-729}{abc - 54}$

$(abc)^{2} - 54(abc) + 729 = 0$ . Differentiating this, we have:

$2(abc) - 54 = 0 \Rightarrow abc = 27$ . Therefore, the answer is 27

ab + 3b = -9
ab = -9 -3b
|abc| = |(-9 -3b)c|
|abc| = |-3(bc + 3c)|
|abc| = |-3(-9)| = |27| = 27

Multiplying all the equations, we get -

$abc(a+3)(b+3)(c+3)=(-9)^3$ $f(a,b,c)=(a^2+3a)(b^2+3b)(c^2+3c)=(-9)^3$

Clearly, $f(a,b,c)$ becomes maximum when $(a^2+3a)=(b^2+3b)=(c^2+3c) ( = (x^2 + 3x))$

Hence $(x^2 + 3x)^3 = (-9)^3 \Rightarrow x^2 + 3x + 9 = 0$

Solving for $x$ gives -

$x = 3\omega, 3\omega^2$ where $\omega = \frac{-1+\sqrt{3}i}{2}$

Therefore, $a=b=c=3\omega, 3\omega^2$

So, $abc = 3^3 = 27$

because $\omega^3 = \omega^6 = 1$

first add all three equation i.e

ab+bc+ca+3(a+b+c)+27=0...............(i)

now multiply all equation i.e,

abc{ab+bc+ca+6(a+b+c)+27}= -729

now on putting the value of eq.(i) in above equation.

we'll get,

|abc|(a+b+c)=243 [on neglecting the (-) sign]

the possible value of a,b&c would be

|3 3 3|(3+3+3) = 3^5

so largest possible value of |abc| would be 27 Ans.