Question

If (a/b+c) + (b/c+a) + (c/a+b) = 1, then find the value of (a^2/b+c) + (b^2/c+a) + (c^2/a+b)

Answer 1

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Answer 2

Answer:

0

Step-by-step explanation:

$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} - 1 = 0$ ⇒ $\frac{a^{3}+b^{3}+c^{3}+abc}{(a+b)(a+c)(b+c)} = 0$

Therefore,

$\frac{a^{2} }{b+c}+\frac{b^{2} }{c+a}+\frac{c^{2} }{a+b}$ = $\frac{a^{4}+a^{3}b+ab^{3}+b^{4}}{y} and so on = 0$ $= (a+b+c)(a^{3}+b^{3}+abc+c^{3})$=0