MCQ
$a + b + c = 0$ then $=\frac{1}{\text{b}^{2}+\text{c}^{2}-\text{a}^{2}}+\frac{1}{\text{c}^{2}+\text{a}^{2}-{\text{b}}^{2}}+\frac{1}{\text{a}^{2}+\text{b}^{2}-\text{c}^{2}}$ is equal to:
- A$3$
- B$6$
- C$1$
- ✓$0$
✓
Answer
Correct option: D.
$0$
D. $0$
Solution:
Given $a + b + c = 0$
$⇒ b + c = -a$
Squaring on both sides
$⇒ b^2 + c^2 + 2bc = a^2$
$⇒ b^2 + c^2 - a^2 = -2bc$
Similarly $c^2 + a^2- b^2 = -2ac$
Similarly $a^2 + b^2 - c^2= -2ab$
⇒ On substituting these values the equation becomes $\frac{-1}{2}\big(\frac{1}{\text{bc}}+\frac{1}{\text{ac}}+\frac{1}{\text{ab}}\Big)$
$\Rightarrow\frac{{-1}}{{2}{\text{abc}}}(\text{a+b+c}) = 0$
Solution:
Given $a + b + c = 0$
$⇒ b + c = -a$
Squaring on both sides
$⇒ b^2 + c^2 + 2bc = a^2$
$⇒ b^2 + c^2 - a^2 = -2bc$
Similarly $c^2 + a^2- b^2 = -2ac$
Similarly $a^2 + b^2 - c^2= -2ab$
⇒ On substituting these values the equation becomes $\frac{-1}{2}\big(\frac{1}{\text{bc}}+\frac{1}{\text{ac}}+\frac{1}{\text{ab}}\Big)$
$\Rightarrow\frac{{-1}}{{2}{\text{abc}}}(\text{a+b+c}) = 0$
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