Question

If p = a(b – c), q = b(c – a), r = c(a – b) then $(\frac{p}{a}{)}^3+(\frac{q}{b}{)}^3+(\frac{r}{c}{)}^3$ is equal to

• A. $\frac{pqr}{3abc}$
  
• B. $\frac{pqr}{abc}$
  
• C. $\frac{abc}{pqr}$
• D. $\frac{3pqr}{abc}$
  

Answer 1

The correct option is D $\frac{3pqr}{abc}$


Given: p = a(b – c), q = b(c – a), r = c(a – b)
$\therefore \frac{p}{a}+\frac{q}{b}+\frac{r}{c}=\frac{a(b-c)}{a}+\frac{b(c-a)}{b}+\frac{c(a-b)}{c}$
$=b-c+c-a+a-b$
= 0
Now, $x^3+y^3+z^3=3xyz$ if $(x+y+z)=0$.
$\therefore (\frac{p}{a}{)}^3+(\frac{q}{b}{)}^3+(\frac{r}{c}{)}^3=3\times \frac{p}{a}\times \frac{q}{b}\times \frac{r}{c}$
$=\frac{3pqr}{abc}$
Hence, the correct answer is option (4).