Given, (54√2x3+24√3y3) ÷(√18x+√12y) =Ax2+By2+Cxy We know that, a3+b3=(a+b)(a2+b2−ab) Then, [(3√2x)3+(2√3y)3] =[3√2x+2√3y] [18x2+12y2−6√6xy] So, compare the equations
(3√2x+2√3y)(18x2+12y2−6√6xy)
(3√2x+2√3y)
=Ax2+By2+Cxy ⇒(18x2+12y2−6√6xy)=Ax2+By2+Cxy Comparing, we get A=18,B=12 and C=−6√6 So, A2−(B2+C2) =(18)2−[122+(−6√6)2] =324−360=−36 A=18,B=12 and C=−6√6 So, A2−(B2+C2)