Question

p^2q-(q^2p-rq-(qr-(p^2q+q^2r)+(q^3-p^3)))+p^3-q^3-r^3 if p=1q=-1r=2 find the value

Answer 1

2p3q5r=p2q3r52p3q5r=p2q3r5

only possible when p=2,q=3,r=5p=2,q=3,r=5

so p+q+r=2+3+5=10p+q+r=2+3+5=10

For those who like detailed answer :

2p3q5rp2q3r5=12p3q5rp2q3r5=1

2pp2.3qq3.5rr5=12pp2.3qq3.5rr5=1

Each fraction needs to be equal to 1 because considering any other value will give you only a relation among fractions.

2pp2=12pp2=1

2p=p22p=p2

Taking log both sides,

2logp=plog22log⁡p=plog⁡2

logpp=log22log⁡pp=log⁡22

p1p=212p1p=212

By comparing p=2p=2, Similarly, q=3q=3 and r=5r=5

p+q+r=10

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