Answers

The Brainliest Answer!
  • Brainly User
2014-03-25T17:17:49+05:30
Qa-b=0
a =\frac{b}{q}
abc= \frac{b^2c}{q}=p
b^2c=pq
b^2= \frac{p}{qc}
similarly
a^2= \frac{pq}{c}
therefore let x be distance from origin
and let d=x^2
d=a^2+b^2+c^2= \frac{pq}{c}+ \frac{p}{qc}+c^2
d'=2c- \frac{pq}{c^2}- \frac{p}{qc^2}
if d'=0 then c^3= \frac{p(q^2+1)}{2q}
and u will also find that
d"(  \sqrt[3]{\frac{p(q^2+1)}{2q} } ) >0
therefore
d= x^{2} =3(\sqrt[3]{\frac{p(q^2+1)}{2q}})^2
and so finally
x= \sqrt{3} \sqrt[3]{\frac{p(q^2+1)}{2q}}






5 5 5
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thanks guys