Question

If a+b+c=0, then write the value of a2/bc+b2/ca+c2/ab

Answer 1

Answer:

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

Explanation:

Given a+b+c=0 ----(1)

we know the algebraic identity:

$\boxed { a^{3}+b^{3}+c^{3}\\=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)+3abc}$

Value of $\frac{a^{2}}{bc}+\frac{b^{2}}{ca}+\frac{c^{2}}{ab}$

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

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

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

= $\frac{0\times(a^{2}+b^{2}+c^{2}-ab-bc-ca)+3abc}{abc}$

/* from (1) */

= $\frac{3abc}{abc}$

= $3$

Therefore,

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

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Answer 2

Step-by-step explanation:

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