Question

Lambda^6 - 3[(lambda)^2] -2=0

Answer 1

f(λ) =λ⁶ −3λ²−2 = 0

Let, λ² =z

So; f(z) = z³–3z–2=0

So,

z₁,₂,₃=ωₖ+ωₖ² (using the cubic formula)

where, ωₖ are three cube roots of one

Therefore,

z₁,₂,₃= 2, –1 , –1

Therefore, the six solutions are :

λ₁ = √2

λ₂ = −√2

λ₃ = √(−1) ≡ i

λ₄ = −√(−1) ≡ −i

λ₅= √(−1) ≡ i

λ₆ = −√(−1) ≡−i