Question

If \( \alpha, \beta, \gamma \) are the roots of \( x^{3} + 4x + 1 = 0 \), then the equation whose roots are \[ \frac{\alpha^{2}}{\beta+\gamma}, \quad \frac{\beta^{2}}{\gamma+\alpha}, \quad \frac{\gamma^{2}}{\alpha+\beta} \] is

• A. $x^{3}-4x-1=0$
• B. $x^{3}-4x+1=0$
• C. $x^{3}+4x-1=0$
• D. $x^{3}+4x+1=0$

Hint:

To negate the roots of an equation, replace $x$ with $-x$.

Answer 1

The Correct Option is C

Step 1: Relationships
For $x^3 + 4x + 1 = 0$, $\alpha + \beta + \gamma = 0$, $\alpha\beta + \beta\gamma + \gamma\alpha = 4$, and $\alpha\beta\gamma = -1$.
Step 2: Simplify Roots
Since $\alpha + \beta + \gamma = 0$, then $\beta + \gamma = -\alpha$. The first root is $\frac{\alpha^2}{-\alpha} = -\alpha$. Similarly, the other roots are $-\beta$ and $-\gamma$.
Step 3: Formation of Equation
To find the equation with roots $-\alpha, -\beta, -\gamma$, replace $x$ with $-x$ in the original equation: $(-x)^3 + 4(-x) + 1 = 0 \Rightarrow -x^3 - 4x + 1 = 0 \Rightarrow x^3 + 4x - 1 = 0$.
Final Answer: (c)