SSC CGL 17 Aug 2021 Shift 1 Paper

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Question : 75 of 100

Marks: +1, -0

If

(54 √{2} x^{3}+24 √{3} y^{3}) ÷

(√{18} x+√{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:

Given,

(54 √{2} x^{3}+24 √{3} y^{3})

÷(√{18} x+√{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 √{2} x)^{3}+(2 √{3} y)^{3}]

=[3 √{2} x+2 √{3} y]

\;\; [18 x^{2}+12 y^{2}-6 √{6} x y]

So, compare the equations

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

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

⇒(18 x^{2}+12 y^{2}-6 √{6} x y)

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

Comparing, we get

A=18, B=12

and

C=-6 √{6}

So,

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

=(18)^{2}-[12^{2}+(-6 √{6})^{2}]

=324-360=-36

A=18, B=12

and

C=-6 √{6}

So,

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

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