Question

IF P+Q+R=0 THEN SHOW THAT P3+Q3+R3=3 PQR

Answer 1

P+Q+R = 0
=> P+Q= -R (equation 1)

cubing both sides

=> (P+Q)³ = (-R)³

=> P³+Q³+ 3PQ(P+Q) = -R³
${(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)$

=> P³+Q³+3PQ (-R) = -R³. (from equation 1)


=> P³+Q³-3PQR = -R³

=> P³+Q³+R³ = 3PQR




PROVED




hope this helps

Answer 2

Answer:

$P^3+Q^3+R^3=0$

Step-by-step explanation:

It is given that

P+Q+R=0

We have to prove that : $P^3+Q^3+R^3=0$

We know that

$P^3+Q^3+R^3-3PQR=(P+Q+R)(P^2+Q^2+R^2-PQ-QR-PR)$

$P^3+Q^3+R^3-3PQR=0\times (P^2+Q^2+R^2-PQ-QR-PR)$

$P^3+Q^3+R^3-3PQR=0$

$P^3+Q^3+R^3=3PQR$

Hence proved.