Question
If $a + b + c = 0$, then find the value of $\frac{(a^2+b^2+c^2)^2}{a^2b^2+b^2c^2+c^2a^2}$
This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
4
Simplifying Algebraic Expression Given $a+b+c=0$
We need to find the value of $\frac{(a^2+b^2+c^2)^2}{a^2b^2+b^2c^2+c^2a^2}$ given the condition $a + b + c = 0$.
Mathematical Derivation Steps
- Use the identity $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$. Since $a + b + c = 0$, we have $0^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$. This simplifies to $a^2 + b^2 + c^2 = -2(ab + bc + ca)$.
- Square both sides of the result from step 1: $(a^2 + b^2 + c^2)^2 = (-2(ab + bc + ca))^2$. $(a^2 + b^2 + c^2)^2 = 4(ab + bc + ca)^2$.
- Use the identity $(ab + bc + ca)^2 = a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a + b + c)$. Given $a + b + c = 0$, this identity simplifies to $(ab + bc + ca)^2 = a^2b^2 + b^2c^2 + c^2a^2$.
- Substitute the result from step 3 into the equation from step 2: $(a^2 + b^2 + c^2)^2 = 4(a^2b^2 + b^2c^2 + c^2a^2)$.
- Substitute this relationship into the expression to be evaluated: $\frac{(a^2+b^2+c^2)^2}{a^2b^2+b^2c^2+c^2a^2} = \frac{4(a^2b^2+b^2c^2+c^2a^2)}{a^2b^2+b^2c^2+c^2a^2}$. Assuming $a^2b^2+b^2c^2+c^2a^2 \neq 0$, the expression simplifies to 4.
The final value of the expression is 4.
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