Question

if a+b+c =0 , then a^2/bc+b^2/ca+c^2/ab=​

Answer 1

Question :--- if a+b+c = 0 , Find the value of (a²/bc) + (b²/ca) + (c²/ab) = ?

Formula used :---

→ a³ + b³ + c³ = (a+b+c)(a² + b² + c² -ab - bc - ca) + 3abc ...

Solution :----

→ (a²/bc) + (b²/ca) + (c²/ab)

Multiplying each parts with a , b and c respectively , we get,

→ (a³/abc) + (b³/abc) + (c³/abc)

Taking LCM now,

→ (a³+b³+b³)/abc

Putting value of (a³+b³+c³) From above Told Formula we get, now,

→ (a+b+c)(a² + b² + c² -ab - bc - ca) + 3abc / abc

Now, putting given value of (a+b+c=0) in Numerator we get,

→ 0 + 3abc/abc

→ 3abc/abc

→ 3 (Ans).

Hence, the value of (a²/bc) + (b²/ca) + (c²/ab) is 3.

Answer 2

$\huge{\bf{Solution:-}}$

Multiplying each parts with a , b and c respectively , we get,

→ (a³/abc) + (b³/abc) + (c³/abc)

Taking LCM now,

→ (a³+b³+b³)/abc

Putting value of (a³+b³+c³) From above Told Formula we get, now,

→ (a+b+c)(a² + b² + c² -ab - bc - ca) + 3abc / abc

Now, putting given value of (a+b+c=0) in Numerator we get,

→ 0 + 3abc/abc

→ 3abc/abc

→ 3 (Ans).

Hence, the value of (a²/bc) + (b²/ca) + (c²/ab) is 3.