MCQ
If $a + b + c = 0,$ then $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}=$
- A$0$
- B$1$
- C$-1$
- ✓$3$
✓
Answer
Correct option: D.
$3$
$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
If $a + b + c = 0$, then
$a^3 + b^3 + c^3 - 3abc = 0$
$\Rightarrow a^3 + b^3 + c^3 = 3abc ...(1)$
Now, consider $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}$
Multiplying dividing by $a. b.$ and $c$ in $\frac{\text{a}^2}{\text{bc}}.\frac{\text{b}^2}{\text{ca}}$ and $\frac{\text{c}^2}{\text{ab}}$ respectively. we get
$\frac{\text{a}^3}{\text{abc}}+\frac{\text{b}^3}{\text{bca}}+\frac{\text{c}^3}{\text{cab}}$
$=\frac{\text{a}^3+\text{b}^3+\text{c}^3}{\text{abc}}$
$=\frac{3\text{abc}}{\text{abc}} ....[$From $(1)]$
$=3$
Hence, correct option is $(d).$
If $a + b + c = 0$, then
$a^3 + b^3 + c^3 - 3abc = 0$
$\Rightarrow a^3 + b^3 + c^3 = 3abc ...(1)$
Now, consider $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}$
Multiplying dividing by $a. b.$ and $c$ in $\frac{\text{a}^2}{\text{bc}}.\frac{\text{b}^2}{\text{ca}}$ and $\frac{\text{c}^2}{\text{ab}}$ respectively. we get
$\frac{\text{a}^3}{\text{abc}}+\frac{\text{b}^3}{\text{bca}}+\frac{\text{c}^3}{\text{cab}}$
$=\frac{\text{a}^3+\text{b}^3+\text{c}^3}{\text{abc}}$
$=\frac{3\text{abc}}{\text{abc}} ....[$From $(1)]$
$=3$
Hence, correct option is $(d).$
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