Question

If p q and r are in ap then p q + r cube minus 8 cube is equal to

Answer 1

Answer:

2q=p+r

r=2q-p

r^3=(2q-p)^3

Answer 2

$p^3+r^3-8q^3=-6pqr$

Step-by-step explanation:

p, q and r are in A.P

Therefore ,$q-p=r-q$

Using definition of A.P

$p+r=q+q=2q$

Cubing on both sides then we get

$(p+r)^3=(2q)^3$

$p^3+r^3+3pr(p+r)=8q^3$

Using identity

$(a+b)^3=a^3+b^3+3ab(a+b)$

$p^3+r^3-8q^3+3pr(2q)=0$

$p^3+r^3-8q^3+6prq=0$

$p^3+r^3-8q^3=-6pqr$

$p^3+r^3-8q^3=-6pqr$

#Learns more:

https://brainly.in/question/1042222:answered by Piyush