Question

If a,b,c are all non-zero and a+b+c=0,prove that a²/bc+b²/ca+c²/ab=3

Answer 1

Answer:

a²/bc  + b²/ca  + c²/ab = 3

Step-by-step explanation:

If a b c are all non zero and a+b+c=0 prove that a2/bc+b2/ca+c2/ab=3

a²/bc  + b²/ca  + c²/ab = 3

Multiplying by abc both sides

=> a³  + b³ + c³  = 3abc

=> (a+ b)³ - 3ab(a+b)  + c³  = 3abc

as we know that a + b + c = 0

=> a + b = -c

=> (-c)³ - 3ab(-c) + c³ = 3abc

=> -c³ + 3abc + c³ = 3abc

=> 3abc = 3abc

=> LHS = RHS

QED

Proved

✔️Hope it will help you.✔️

Answer 2

$\huge\sf\blue{Given}$

➝ a,b,c are all non zeros

➝ a+b+c = 0

$\rule{110}1$

$\huge\sf\gray{To\;Prove}$

$\sf\leadsto \dfrac{a^2}{bc}+ \dfrac{b^2}{ca} + \dfrac{c^2}{ab} = 3$

$\rule{110}1$

$\huge\sf\purple{Steps}$

We know that,

If x + y + z = 0, then x³ + y³ + z³ = 3xyz.

$\sf \frac{ {a}^{2} }{bc} + \frac{ {b}^{2} }{ca} + \frac{ {c}^{2} }{ab} \\ \\ \dashrightarrow \sf \frac{a {}^{3} + b {}^{3} + c {}^{3} }{abc} \\ \\ \dashrightarrow \sf \frac{3abc}{abc} \\ \\ \dashrightarrow\sf 3$

Hence proved!

$\sf{\underline{\bullet{\;Extra\:Info}}}$

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)

(a + b)² = a² + b² + 2ab

(a - b)² = a² + b² - 2ab

(x + a)(x + b) = x² + (a + b)x + ab

$\rule{170}3$