Question

Solve for x :
$\frac{x - {a}^{2} }{b + c} + \frac{x - {b}^{2} }{c + a} + \frac{x - {c}^{2} }{a + b} = 4(a + b +c)$
Please answer the question.​

Answer 1

Question:-

Solve for x :

$\sf \large\frac{x - {a}^{2} }{b + c} + \frac{x - {b}^{2} }{c + a} + \frac{x - {c}^{2} }{a + b} = 4(a + b +c).$

Given:-

$\sf \large \frac{x - {a}^{2} }{b + c} + \frac{x - {b}^{2} }{c + a} + \frac{x - {c}^{2} }{a + b} = 4(a + b +c).$

To Find:-

• The value of x.

Solution:-

$\sf \large \frac{x - {a}^{2} }{b + c} + \frac{x - {b}^{2} }{c + a} + \frac{x - {c}^{2} }{a + b} = 4(a + b +c).$

$\sf \large \Rightarrow \bigg \{ \frac{x - a {}^{2} }{b + c} - (b + c + 2a) \bigg \} + \bigg \{ \frac{x - b {}^{2} }{c + a} - (c + a + 2b) \bigg \} + \bigg \{ \frac{x - c {}^{2} }{a + b} - (a + b + 2c) \bigg \}.$

$\sf \large ⇒ \bigg \{ \frac{x - a {}^{2} { - b }^{2} - c {}^{2} - 2ab - 2bc - 2ca }{b + c} \bigg \} + \bigg \{ \frac{x - a {}^{2} - b {}^{2} - c {}^{2} - 2ab - 2bc - 2ca }{c + a} \bigg \} + \bigg \{ \frac{x - a {}^{2} - b { }^{2} - c {}^{2} - 2ab - 2bc - 2ca }{a + b} \bigg \} = 0.$

$\sf \large⇒ (x - a {}^{2} - b {}^{2} - c {}^{2} - 2ab - 2bc - 2ca) \: \bigg \{ \frac{1}{b + c} + \frac{1}{c + a} + \frac{1}{a + b} \bigg \} = 0.$

$\sf \large⇒ - \bigg \{x - (a + b + c) {}^{2} \bigg \} \: \bigg \{ \frac{1}{b + c} + \frac{1}{c + a} + \frac{1}{a + b} \bigg \} = 0.$

$\sf \large Thus, \: x - (a + b + c) {}^{2} = 0 \: and \\ \sf \large \frac{1}{b +c} + \frac{1}{c + a} + \frac{1}{a + b} ≠ 0.$

$\sf \large ⇒ (a + b + c) {}^{2} .$

Answer:-

$\sf \large \therefore x = (a + b + c) {}^{2} .$

Hope you have satisfied. ⚘