Question

if $p = a(b-c) , q = b(c-a) , r = c(a-b) then (p/a)^{3} + (q/b)^{3} + (r/c)^{3}$

Answer 1

Answer:

Given:

• $p=a(b-c)$
• $q = b(c - a)$
• $r = c(a-b)$

Solution:

$\left( \dfrac{p}{a} \right)^3 + \left( \dfrac{q}{b} \right)^3 + \left( \dfrac{r}{c} \right)^3$

$\implies \left( \dfrac{a\!\!\!／(b-c)}{a\!\!\!／} \right)^3 + \left( \dfrac{b\!\!\! ／(c-a)}{b\!\!\!／} \right)^3 + \left( \dfrac{c\!\!\!／(a-b)}{c\!\!\!／} \right)^3$

$\implies (b-c)^3 + (c-a)^3 + (a-b)^3$

Let:

$b- c = x \\ c- a = y \\ a-b = z$

$\implies x + y+z = b-c + c-a + a-b$

$\implies x + y + z = 0$

By identity :

$\therefore x^3 + y^3 + z^3 = 3xyz$

$\implies (b-c)^3 + (c-a)^3 + (a-b)^3 = 3(a-b)(b-c)(c-a)$

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