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 $A^2-\left(B^2+C^2\right)$ ?
Answer & explanation
Correct answer: option 2
(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