If (1029\sqrt3x^3-54\sqrt2y^3)\div(7\sqrt3x-3\sqrt2y) =\A x^2+\sqrt6\B xy+\C y^2, then the value of (3\A-13\B-13/9\C)?

SSC CGL Model Paper 13

Question : 59 of 100

Marks: +1, -0

(1029√3x^3-54√2y^3)÷(7√3x-3√2y)

=\A x^2+√6\B xy+\C y^2,

then the value of

(3\A-13\B-13/9\C)?

Solution:

\A x^2+√6\B xy+\C y^2

=147x^2+18y^2+21√6 xy

A = 147, B = 21, C = 18
Now

3\A-13\B-13/9\C

=3(147)-13(21)-13/9(18)

= 441 - 273 - 26 = 142

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