Question

Let 'a' be a non-zero real number. If the equation whose roots are the squares of the roots of the cubic equation \( x^3 - ax^2 + ax - 1 = 0 \) is identical with this cubic equation, then 'a' =

• A. \( \frac{1}{3} \)
• B. \( 3 \)
• C. \( \frac{1}{2} \)
• D. \( 2 \) Correct Answer

Hint:

If \( \alpha, \beta, \gamma \) are roots of \( P(x) = 0 \), to find an equation whose roots are \( f(\alpha), f(\beta), f(\gamma) \), let \( y = f(x) \). Solve for \(x\) in terms of \(y\), say \( x = g(y) \), and substitute this into \( P(x)=0 \) to get \(P(g(y))=0\). For roots being squares (\(y=x^2\)), \(x=\sqrt{y}\). Isolate radical terms and square to rationalize. If two polynomials are identical, their corresponding coefficients must be proportional (or equal if leading coefficients are equal).

Answer 1

The Correct Option is B

Step 1: Let the given cubic equation be \( P(x) = x^3 - ax^2 + ax - 1 = 0 \).
Let its roots be \( \alpha, \beta, \gamma \).

Step 2: Form the equation whose roots are \( \alpha^2, \beta^2, \gamma^2 \).
Let \( y = x^2 \), so \( x = \sqrt{y} \).
Substitute this into \( P(x)=0 \): \[ (\sqrt{y})^3 - a(\sqrt{y})^2 + a\sqrt{y} - 1 = 0 \] \[ y\sqrt{y} - ay + a\sqrt{y} - 1 = 0 \] Rearrange to isolate terms with \( \sqrt{y} \): \[ \sqrt{y}(y+a) = ay+1 \] Square both sides to eliminate the square root: \[ y(y+a)^2 = (ay+1)^2 \] \[ y(y^2 + 2ay + a^2) = a^2y^2 + 2ay + 1 \] \[ y^3 + 2ay^2 + a^2y = a^2y^2 + 2ay + 1 \] \[ y^3 + (2ay^2 - a^2y^2) + (a^2y - 2ay) - 1 = 0 \] \[ y^3 + (2a-a^2)y^2 + (a^2-2a)y - 1 = 0 \] This is the equation whose roots are \( \alpha^2, \beta^2, \gamma^2 \).
Let this be \( Q(y)=0 \).

Step 3: Compare the new equation with the original equation.
The problem states that \( Q(y)=0 \) (or \( Q(x)=0 \) if we use \(x\) as the variable) is identical to \( P(x)=0 \).
Original equation: \( x^3 - ax^2 + ax - 1 = 0 \) New equation: \( x^3 + (2a-a^2)x^2 + (a^2-2a)x - 1 = 0 \) For these equations to be identical, their corresponding coefficients must be equal (since the coefficient of \(x^3\) is 1 in both).
Comparing coefficient of \(x^2\): \[ -a = 2a-a^2 \] \[ a^2 - 3a = 0 \] \[ a(a-3) = 0 \] Since 'a' is a non-zero real number, we must have \( a-3=0 \), so \( a=3 \).

Step 4: Compare coefficient of \(x\): \[ a = a^2-2a \] \[ a^2 - 3a = 0 \] \[ a(a-3) = 0 \] Again, since \( a \ne 0 \), we have \( a=3 \).
The constant terms (-1 and -1) are already equal.
Both comparisons yield \( a=3 \).
This matches option (2).