Question : If $\left(5 \sqrt{5} x^3-3 \sqrt{3} y^3\right) \div(\sqrt{5} x-\sqrt{3} y)=\left(A x^2+B y^2+C x y\right)$, then the value of $(3 A+B-\sqrt{15} C)$ is:
Option 1: 8
Option 2: 5
Option 3: 3
Option 4: 12
Correct Answer: 3
Solution :
Given: $\left(5 \sqrt{5} x^3-3 \sqrt{3} y^3\right) \div(\sqrt{5} x-\sqrt{3} y)=\left(A x^2+B y^2+C x y\right)$
⇒ $((\sqrt{5} x)^3-(\sqrt{3} y)^3) \div(\sqrt{5} x-\sqrt{3} y)=\left(A x^2+B y^2+C x y\right)$
⇒ $((\sqrt{5} x-\sqrt{3} y)((\sqrt{5} x)^2+(\sqrt{3} y)^2+(\sqrt{5} x)×(\sqrt{3} y)) \div(\sqrt{5} x-\sqrt{3} y)=\left(A x^2+B y^2+C x y\right)$
⇒ $(\sqrt{5} x)^2+(\sqrt{3} y)^2+(\sqrt{5} x)×(\sqrt{3} y)=\left(A x^2+B y^2+C x y\right)$
⇒ $5x^2+3y^2+\sqrt{15} x y=\left(A x^2+B y^2+C x y\right)$
⇒ $A = 5$, $B = 3$, and $C = \sqrt{15}$
So, $(3 A+B-\sqrt{15} C)$
= $3×5 + 3 - \sqrt{15}×\sqrt{15}$
= $15+3-15$
= $3$
Hence, the correct answer is 3.
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