Question

If a, b, c are all non zero and a + b+ c = 0, prove that a²/bc+b²/ca+²/=3

Answer 1

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

Answer:

We know that,

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

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

Hence, it is proved.

$\rule{300}{2}$

$\large{\underline{\underline{\mathfrak{\heartsuit \: Algebraic \: Identity : \: \heartsuit}}}}$

• (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

Answer 2

Correct question:

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

Answer

Let's simplify the given equation first:

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

We know an identity:

When: a + b + c = 0;

${a}^{3} + {b}^{3} + {c}^{3} = 3abc$

Applying the same here:

$\frac{ {a}^{3} + {b}^{3} + {c}^{3} }{abc} \\ \\ = > \frac{3abc}{abc} \\ \\ = > 3$