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SSC CGL 17 Aug 2021 Shift 1 Paper

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Question : 75 of 100

Marks: +1, -0

If

(54 \sqrt{2} x^{3}+24 \sqrt{3} y^{3}) \div

(\sqrt{18} x+\sqrt{12} y)=A x^{2}+B y^{2}+C x y

, then what is the value of

A^{2}-(B^{2}+C^{2})

?

12
$-36$
$-24$
24

Solution: 👈: Video Solution

Given,

(54 \sqrt{2} x^{3}+24 \sqrt{3} y^{3})

\div(\sqrt{18} x+\sqrt{12} y)

=A x^{2}+B y^{2}+C x y

We know that,

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

Then,

\;\;[(3 \sqrt{2} x)^{3}+(2 \sqrt{3} y)^{3}]

=[3 \sqrt{2} x+2 \sqrt{3} y]

\;\;[18 x^{2}+12 y^{2}-6 \sqrt{6} x y]

So, compare the equations

\;\frac{(3 \sqrt{2} x+2 \sqrt{3} y)(18 x^{2}+12 y^{2}-6 \sqrt{6} x y)}{(3 \sqrt{2} x+2 \sqrt{3} y)}

\;\;=A x^{2}+B y^{2}+C x y

\Rightarrow(18 x^{2}+12 y^{2}-6 \sqrt{6} x y)

=A x^{2}+B y^{2}+C x y

Comparing, we get

A=18, B=12

and

C=-6 \sqrt{6}

So,

A^{2}-(B^{2}+C^{2})

=(18)^{2}-[12^{2}+(-6 \sqrt{6})^{2}]

=324-360=-36

A=18, B=12

and

C=-6 \sqrt{6}

So,

A^{2}-(B^{2}+C^{2})

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