If 2 √(2) x³ - 3 √(3) y³ = ( √(2x)-√(3y) ) (Ax² - Bxy +Cy²), then the value of √() (A² + B² + C²) is:

If 2 √(2) x³ - 3 √(3) y³ = ( √(2x)-√(3y) ) (Ax² - Bxy…

If 2 $\sqrt{2}$ x³ - 3 $\sqrt{3}$ y³ = ( $\sqrt{2x}-\sqrt{3y}$ ) (Ax² - Bxy +Cy²), then the value of $\sqrt{}$ (A² + B² + C²) is:

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Correct answer: A

Explanation

The correct answer is option 1 i.e. $\sqrt{19}$ .

x³ - y³ = (x - y)(x²+ xy + y²)

2 $\sqrt{2}$ x³ - 3 $\sqrt{3}$ y³ = ( $\sqrt{2x}-\sqrt{3y}$ ) (Ax² - Bxy +Cy²)

⇒ (2√2x³ - 3√3y³)/(√(2x) - √(3y)) = Ax² - Bxy + Cy²

⇒ $\frac{(\sqrt2x - \sqrt3y)(2x^2 + 3y^2 - (\sqrt6xy)}{\sqrt2x - \sqrt3y}$ = Ax² - Bxy + Cy²

= (2x²- √6xy + 3y²) = = Ax² - Bxy + Cy²

On comparing with Ax² - Bxy + Cy² -

A = 2

B = √6

C = 3

⇒ √(A² + B² + C²) = √{2² + (√6)² + 3²} = √(4 + 6 + 9) = √19