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If ∣1xx2x21xxx21∣=(xλ−1)μ\begin{vmatrix} 1 & x & x^2 \\ x^2 & 1 & x \\ x & x^2 & 1 \end{vmatrix} = (x^{\lambda}-1)^{\mu} then the value of λ\lambda and μ\mu are given by

Solution

✅ Correct Option: 3

Expanding along the first row gives 1(1−x3)−x(x2−x2)+x2(x4−x)1(1-x^3) - x(x^2-x^2) + x^2(x^4-x).

This simplifies to 1−x3+x6−x3=x6−2x3+1=(x3−1)21-x^3+x^6-x^3 = x^6-2x^3+1 = (x^3-1)^2.

Comparing with (xλ−1)μ(x^{\lambda}-1)^{\mu} gives λ=3\lambda = 3 and μ=2\mu = 2.

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