Option 3 : 2√2

Electric charges and coulomb's law (Basic)

48838

10 Questions
10 Marks
10 Mins

**GIVEN:**

√(p^{3} + q^{3} + r^{3}) = 6, (p + q + r)^{2} = 36 and 2(pq + qr + rp) = 22

**CONCEPT:**

Algebraic identity.

**FORMULA USED:**

(p^{3} + q^{3} + r^{3}) = (p + q + r)(p^{2} + q^{2} + r^{2} – pq – qr – rp) + 3pqr

**CALCULATION:**

Given that,

(p + q + r)^{2} = 36 …….(i)

⇒ p^{2} + q^{2} + r^{2} + 2(pq + qr + rp) = 36

p^{2} + q^{2} + r^{2} + 22 = 36

p^{2} + q^{2} + r^{2} = 14 ……..(ii)

Also given that,

√(p^{3} + q^{3} + r^{3}) = 6

⇒ (p^{3} + q^{3} + r^{3}) = 36

Now,

(p + q + r)(p^{2} + q^{2} + r^{2} – pq – qr – rp) + 3pqr = 36

From equation (i) and (ii):

6(14 – 11) + 3pqr = 36

⇒ 3pqr = 18

⇒ pqr = 6

Now

√(4pqr/3)

⇒ √(4×6/3) = 2√2