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Maths Question 12 – AP-EAMCET 2026

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If α,β,γ are roots of the equation x3+2x2−x−2=0, then α6+β6+γ6=

Identify the roots of the given cubic equation x3+2x2−x−2=0. Look for simple integer roots first, which can help factor the polynomial.

Step 1: Find the roots of the cubic equation✦ Active

The given cubic equation is x3+2x2−x−2=0. We can find the roots by testing simple integer values for x.

For x=1, substitute into the equation: P(1)=(1)3+2(1)2−(1)−2=1+2−1−2=0. Thus, x=1 is a root.

Since x=1 is a root, (x−1) is a factor of the polynomial. We can use synthetic division to find the other factors:

112−1−21321320

The quotient is x2+3x+2. Now, factor this quadratic equation: x2+3x+2=0⟹(x+1)(x+2)=0.

So, the remaining roots are x=−1 and x=−2. Therefore, the roots of the equation are α=1, β=−1, and γ=−2 (the assignment to α,β,γ is interchangeable).

💡 Teacher's Secret Hint

Always check for simple integer roots like ±1,±2 first. They often simplify the factoring process significantly. If P(a)=0, then (x−a) is a factor.

Step 2: Calculate the sum of the sixth powers of the roots○ Expand

We need to calculate the value of α6+β6+γ6.

Substitute the roots we found into the expression:

α6=(1)6=1 β6=(−1)6=1 γ6=(−2)6=64

Now, sum these values:

α6+β6+γ6=1+1+64=66
💡 Teacher's Secret Hint

When raising negative numbers to powers, remember that an even exponent results in a positive value (e.g., (−2)6=64), while an odd exponent retains the negative sign (e.g., (−2)3=−8). Pay attention to the signs.

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