Question

if p2+4q2+9r2=2pq+6qr+3pr then prove that p3+8q3+27r3=18pqr​

Answer 1

Answer:

12+3++473448248₱7-74+2₱38₱883836(48&)9₱+₱8₱23

Answer 2

Answer:

Step-by-step explanation: Given that p²+4q²+9r² = 2pq+6gr + 3pr...(1)

We know that

a³ + b³ + c³ −3abc

= (a+b+c)(a² + b² + c² -ab-bc-ca) ....(2)

Here, a = p,b=2q and c=3r.

Then using the above identity (2) we have,

p³ +8q3+27r³-3xpx2qx3r

= (p+2q+3r) (p²+4q²+9r2-2pq-6qr-3pr)

→ p³ +8q³ +27r³ – 18pqr

= (p+2q+3r)(p²+4q²+9r² - 2pq-6qr-3pr)

→ p³+8q³ +27r³ - 18pqr = (p+2q+3r)x0 [p²+4q²+9r² = 2pq+6qr+3pr]

p3+8q3+27r3 - 18pqr=0

→p3+8q3+27r3

= 18pqr

HENCE PROVED!