If $\left(56 \sqrt{7} x^3-2 \sqrt{2} y^3\right) \div(2 \sqrt{7} x-\sqrt{2} y)=\mathrm{A} x^2+B y^2-C x y$, then find the value of $A+B-\sqrt{14} C$.

58

( a - b ) = \(\frac{a^3 - b^3}{a^2 + b^2 + ab }\)

If $\left(56 \sqrt{7} x^3-2 \sqrt{2} y^3\right) \div(2 \sqrt{7} x-\sqrt{2} y)=\mathrm{A} x^2+B y^2-C x y$,]

On comparing the above equation with ( a - b ) = \(\frac{a^3 - b^3}{a^2 + b^2 + ab }\) we can conclude that ,

a = ($2 \sqrt{7}$)² = 28

b = ($\sqrt{2}$)² = 2

c = -$2 \sqrt{7}$ × $\sqrt{2}$ = -2\(\sqrt {14}\)

So, put them in  $A+B-\sqrt{14} C$ = 28 + 2 -(- 2\(\sqrt {14}\) × $\sqrt{14}$

= 30 + 28 = 58