If $\left(54 \sqrt{2} x^3+24 \sqrt{3} y^3\right) \div(\sqrt{18} x+\sqrt{12} y)=A x^2+B y^2+C x y$, then what is the value of $A^2-\left(B^2+C^2\right)$ ?

-36

(54√2 x³ + 24√3 y³) ÷ (√18 x + √12 y) = Ax² + By² + Cxy

a³ + b³ = (a + b)(a² + b² – ab)

By comparing the values of given equation with the formula we get the values of A, B and C as given below,

where A = (18), B = (12) and C = (-6√6)

Now put them in  $A^2-\left(B^2+C^2\right)$ 

The value of  A² – (B² + C²) = (18)² – (12)² + (6√6)²

= 324 – (144 + 216)

= 324 – 360

=  -36