If alpha, beta, gamma are the roots of the equation x^3 + 3x^2 - x - 3 = 0, then (1+alpha^2)(1+beta^2)(1+gamma^2) =

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q12 – AP-EAMCET 2026

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Home AP-EAMCET 2026 (12th May 2026 Shift 1)MathsQuestion #12

α, β, γ

are the roots of the equation

x^{3} + 3 x^{2} - x - 3 = 0

, then

(1 + α^{2}) (1 + β^{2}) (1 + γ^{2}) =

Clue 1 of 3: Polynomial Roots and Values

The product $(1 + α^{2}) (1 + β^{2}) (1 + γ^{2})$ can be related to the polynomial function $P (x)$ evaluated at specific complex numbers.

🧠 Full Solution Path

Step 1: Relate the expression to the polynomial✦ Active

Let the given polynomial be $P (x) = x^{3} + 3 x^{2} - x - 3$ . Since $α, β, γ$ are the roots of $P (x) = 0$ , we can write $P (x) = (x - α) (x - β) (x - γ)$ . We need to find the value of the expression $(1 + α^{2}) (1 + β^{2}) (1 + γ^{2})$ . This expression can be related to the polynomial evaluated at complex numbers.

(1 + α^{2}) (1 + β^{2}) (1 + γ^{2}) = (- (- 1) - α^{2}) (- (- 1) - β^{2}) (- (- 1) - γ^{2})

= - (i^{2} - α^{2}) (i^{2} - β^{2}) (i^{2} - γ^{2}) (This is incorrect formulation, better use product of (i - r o o t) (i + r o o t))

We evaluate

P (i)

and

P (- i)

P (i) = (i - α) (i - β) (i - γ)

P (- i) = (- i - α) (- i - β) (- i - γ) = - (i + α) (i + β) (i + γ)

Now, consider the product

P (i) P (- i)

P (i) P (- i) = (i - α) (i - β) (i - γ) \cdot [- (i + α) (i + β) (i + γ)]

= - [(i - α) (i + α)] [(i - β) (i + β)] [(i - γ) (i + γ)]

= - [i^{2} - α^{2}] [i^{2} - β^{2}] [i^{2} - γ^{2}]

Since

i^{2} = - 1

= - [- 1 - α^{2}] [- 1 - β^{2}] [- 1 - γ^{2}]

= - (- 1)^{3} (1 + α^{2}) (1 + β^{2}) (1 + γ^{2})

= - (- 1) (1 + α^{2}) (1 + β^{2}) (1 + γ^{2})

= (1 + α^{2}) (1 + β^{2}) (1 + γ^{2})

💡 Teacher's Secret Hint

This identity, $(1 + α^{2}) (1 + β^{2}) (1 + γ^{2}) = P (i) P (- i)$ , is a standard result for cubic polynomials and is very useful for expressions involving sums of squares of roots.

Step 2: Evaluate P(i)○ Expand

Substitute $x = i$ into the polynomial $P (x) = x^{3} + 3 x^{2} - x - 3$ .

P (i) = i^{3} + 3 i^{2} - i - 3

Recall that

i^{2} = - 1

and

i^{3} = i \cdot i^{2} = i \cdot (- 1) = - i

P (i) = (- i) + 3 (- 1) - i - 3

P (i) = - i - 3 - i - 3

P (i) = - 6 - 2 i

💡 Teacher's Secret Hint

Be careful with the powers of $i$ . A common mistake is $i^{3} = i$ instead of $- i$ .

Step 3: Evaluate P(-i)○ Expand

Substitute $x = - i$ into the polynomial $P (x) = x^{3} + 3 x^{2} - x - 3$ .

P (- i) = (- i)^{3} + 3 (- i)^{2} - (- i) - 3

Recall that

(- i)^{2} = i^{2} = - 1

and

(- i)^{3} = - (i^{3}) = - (- i) = i

P (- i) = (i) + 3 (- 1) - (- i) - 3

P (- i) = i - 3 + i - 3

P (- i) = - 6 + 2 i

💡 Teacher's Secret Hint

Notice that $P (- i)$ is the complex conjugate of $P (i)$ when the polynomial coefficients are real. This can serve as a quick check for your calculations.

Step 4: Calculate the product P(i)P(-i)○ Expand

Now, multiply the values of $P (i)$ and $P (- i)$ to find the required expression.

(1 + α^{2}) (1 + β^{2}) (1 + γ^{2}) = P (i) P (- i)

P (i) P (- i) = (- 6 - 2 i) (- 6 + 2 i)

This is in the form

(a - b) (a + b) = a^{2} - b^{2}

, where

a = - 6

and

b = 2 i

= (- 6)^{2} - (2 i)^{2}

= 36 - 4 i^{2}

Since

i^{2} = - 1

= 36 - 4 (- 1)

= 36 + 4

= 40

💡 Teacher's Secret Hint

For this particular polynomial, the roots are easy to find by factoring: $x^{3} + 3 x^{2} - x - 3 = x^{2} (x + 3) - 1 (x + 3) = (x^{2} - 1) (x + 3) = (x - 1) (x + 1) (x + 3) = 0$ . The roots are $1, - 1, - 3$ . You can verify the answer by direct substitution: $(1 + 1^{2}) (1 + (- 1)^{2}) (1 + (- 3)^{2}) = (1 + 1) (1 + 1) (1 + 9) = 2 \times 2 \times 10 = 40$ . Both methods yield the same result.

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Maths Question 12 – AP-EAMCET 2026

Clue 1 of 3: Polynomial Roots and Values

🧠 Full Solution Path