Срочноо!!! x^4+1/x^4 при а)x+1/x=3; b)x-1/x=5

Решите задачу:

$\left(x\pm\dfrac1x\right)^4=x^4\pm4x^2+6\pm\dfrac4{x^2}+\dfrac1{x^4}\\ x^4+\dfrac1{x^4}=\left(x\pm\dfrac1x\right)^4\mp4\left(x^2+\dfrac1{x^2}\right)-6\\ x^4+\dfrac1{x^4}=\left(x\pm\dfrac1x\right)^4\mp4\left(x\pm\dfrac1{x}\right)^2+8-6\\ a)\;x^4+\dfrac1{x^4}=3^4-4\cdot3^2+8-6=47\\ b)\;x^4+\dfrac1{x^4}=5^4+4\cdot5^2+8-6=727$