Question

is the centroid of a triangle is formed by the points A (p,q),B(q,r) and C(r,p) is at the origin,then find the value of $p^{3} + q^{3}+ r^{3}$

Answer 1

Formula we will be using:
i) Centroid of a triangle when the points are (a,b), (c,d) and (e,f), is given by
Centroid = $( \frac{a+c+e}{3}, \frac{b+d+f}{3})$
ii) $(a+b+c)^3=a^3+b^3+c^3+3ab(a+b)+3c(a+b)(a+b+c)$
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Given points are A(p,q), B(q,r) and C(r,p)
Then, centroid =$( \frac{p+q+r}{3}, \frac{q+r+p}{3})$

Also, given that the centroid is at the origin (0,0)

According to question,
$( \frac{p+q+r}{3}, \frac{q+r+p}{3})=(0,0)$
Comparing both sides:
⇒$\frac{p+q+r}{3}=0 \text{ and } \frac{q+r+p}{3}=0$
Multiplying both sides by 3:
⇒$p+q+r=0\text{ and } q+r+p=0$
⇒p+q+r=0
⇒p+q = -r.....(i)

Therefore,
[tex](p+q+r)^3=0^3 \\ (p+q+r)^3=0 [/tex]
Applying $(a+b+c)^3=a^3+b^3+c^3+3ab(a+b)+3c(a+b)(a+b+c)$:
$p^3+q^3+r^3+3pq(p+q)+3r(p+q)(p+q+r)=0$
Plugging in (p+q+r)=0 and (p+q)=-r:
[tex]p^3+q^3+r^3+3pq(-r)+3r(-r)(0)=0 \\ p^3+q^3+r^3-3pqr+0=0 \\ \text{ Adding 3pqr to both sides} \\ p^3+q^3+r^3=3pqr [/tex]

Answer:
$p^3+q^3+r^3=3pqr$