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 what is the value of $(3 A-B-\sqrt{15} C)$?
Option 1: –3
Option 2: –5
Option 3: 8
Option 4: 12
Correct Answer: –3
Solution :
Given: $\left(5 \sqrt{5} x^3-3 \sqrt{3} y^3\right) \div(\sqrt{5} x-\sqrt{3} y)=\left(A x^2+B y^2+C x y\right)$
⇒ $\frac{(5 \sqrt{5} x^3-3 \sqrt{3} y^3)}{(\sqrt{5} x-\sqrt{3} y)}=\left(A x^2+B y^2+C x y\right)$
⇒ $\frac{(x\sqrt{5})^3-(y\sqrt{3})^3)}{(\sqrt{5} x-\sqrt{3} y)}=(A x^2+B y^2+C x y)$
⇒ $\frac{(\sqrt{5}x-\sqrt{3}y)(5x^2+\sqrt{15}xy+3y^2)}{(\sqrt{5} x-\sqrt{3} y)}=(A x^2+B y^2+C x y)$
⇒ $5x^2+\sqrt{15}xy+3y^2=A x^2+B y^2+C x y$
On comparing,
⇒ $A=5$
⇒ $B=3$
⇒ $C=\sqrt{15}$
Now, $(3 A-B-\sqrt{15} C)$
Putting the values, we get:
= $(3\times5-3-\sqrt{15}\times\sqrt{15})$
= $(15-3-15)$
= –3
Hence, the correct answer is –3.
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