Don't ask about the kangaroo

$0={ a }^{ 2 }(b+c)-{ b }^{ 2 }(a+c)$

$=ab(a-b)+{ (a }^{ 2 }-{ b }^{ 2 })c$

$=(a-b)(ab+ac+bc)$ but $a\neq b$ ,

then $ab+ac+bc =0$ .

Multiplying by $a-c$

$0=(a-c)(ab+ac+bc)$

$0=ac(a-c)+ ({ a }^2-{ c}^2)b$

$0=a^2(b+c)-{ c }^2(a+b)$

then ${ c }^2(a+b)= a^2(b+c)={ b }^{ 2 }(a+c)=\boxed{2}$

$a^2(b+c)=b^2(a+c)=2$ $a^2(b+c)=b^2(a+c)$ $c(a^2-b^2)=ab(b-a)$ $c(a+b)(a-b)=-ab(a-b)$ $(c(a+b))^2=(-ab)^2$ $c^2(a+b)= \frac {a^2b^2} {a+b}$

From the original equations: $c=\frac {2} {a^2}-b=\frac {2} {b^2}-a$ $a-b=\frac {2} {b^2}-\frac {2} {a^2}$

$a-b=2 \frac {(a+b)(a-b)} {a^2b^2}$ $\frac {a^2b^2} {a+b}=\boxed {c^2(a+b)=2}$

Since

(a^2)( b + c ) = (b^2)( a + c )

Then

( a - b )( a b + a c + b c ) = 0 , but a, b are not equal, then

a b + a c + b c = 0

c( a + b ) = - a b

Multiply by c

Then

(c^2)( a + b ) = - a b c ......................... (1)

Since

(a^2)( b + c ) = (b^2)( a + c ) = 2

Then

a + c = 2/b^2

b + c = 2/a^2

eliminating c

Then

(2/b^2) - a = (2/a^2) - b

Simplifying , then we get

a + b = (a^2)(b^2)/2

multiply by c^2

Then

(c^2)( a + b ) =(a^2)(b^2)(c^2)/2 ...............(2)

From (1) , (2) we get

a b c = -2

substituting in (1) we get

(c^2)( a + b ) = 2