x2−4√2kx+2e4lnk−1=0 or, x2−4√2kx+2k4−1=0 α+β=4√2k and α⋅β=2k4−1 Squaring both sides, we get (α+β)2=(4√2k)2⇒α2+β2+2αβ=32k2 66+2αβ=32k2 66+2(2k4−1)=32k2 66+4k4−2=32k2⇒4k4−32k2+64=0 or, k4−8k2+16=0⇒(k2)2−8k2+16=0 ⇒(k2−4)(k2−4)=0⇒k2=4,k2=4 ⇒k=±2 Now, α3+β3=(α+β)(α2+β2−αβ) ∴α3+β3=(4√2k)[66−(2k4−1)] Putting k=−2,(k=+2 cannot be taken because it does not satisfy the above equation) ∴α3+β3=(4√2(−2))[66−2(−2)4−1] α3+β3=(−8√2)(66−32+1)=(−8√2) ∴α3+β3=−280√2