Question
If a + b + c = 0, then write the value of $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}.$
✓
Answer
We have to find the value of $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}$
Given a + b + c = 0
Using identity a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
Put a + b + c = 0
a3 + b3 + c3 - 3abc = (0)(a2 + b2 + c2 - ab - bc - ca)
a3 + b3 + c3 - 3abc = 0
a3 + b3 + c3 = 3abc
$\frac{\text{a}^3}{\text{abc}}+\frac{\text{b}^3}{\text{abc}}+\frac{\text{c}^3}{\text{abc}}=3$
$\frac{\text{a}\times\text{a}\times\text{a}}{\text{abc}}+\frac{\text{b}\times\text{b}\times\text{b}}{\text{abc}}+\frac{\text{c}\times\text{c}\times\text{c}}{\text{abc}}=3$
$\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}=3$
Hence the value of $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}$ is 3.
Given a + b + c = 0
Using identity a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
Put a + b + c = 0
a3 + b3 + c3 - 3abc = (0)(a2 + b2 + c2 - ab - bc - ca)
a3 + b3 + c3 - 3abc = 0
a3 + b3 + c3 = 3abc
$\frac{\text{a}^3}{\text{abc}}+\frac{\text{b}^3}{\text{abc}}+\frac{\text{c}^3}{\text{abc}}=3$
$\frac{\text{a}\times\text{a}\times\text{a}}{\text{abc}}+\frac{\text{b}\times\text{b}\times\text{b}}{\text{abc}}+\frac{\text{c}\times\text{c}\times\text{c}}{\text{abc}}=3$
$\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}=3$
Hence the value of $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}$ is 3.
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