If 6 √(6p³) + 2 √(2q³) =( √(6p)+√(2q) )(Sp² + Mq² - Npq), ten the positive value of √(S²+M²+2N²) is:

If 6 √(6p³) + 2 √(2q³) =( √(6p)+√(2q) )(Sp² + Mq² - Npq)…

If 6 $\sqrt{6p^3}$ + 2 $\sqrt{2q^3}$ =( $\sqrt{6p}+\sqrt{2q}$ )(Sp² + Mq² - Npq), ten the positive value of $\sqrt{S^2+M^2+2N^2}$ is:

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Correct answer: A

Explanation

The correct answer is option 1 i.e. 8

6(6)^½ p³ + 2(2)^½q³ = ((6)^½p + (2)^½q) [(6)^½p)² + ((2)^½q)² - (6)^½× (2)^½pq)

So, S = 6

M = 12

N = (12)^½

Now, (S² + M² + 2N²)^½ = 6² + 2² + 2 × [(12)^½]²

⇒ [36 + 4 + 2 × 12 ]^½

⇒ (64)^½ = 8.