已知a满足等式a的平方减a减1等于0,求代数式a的8次方加｛7乘a的负4次方｝的值

a² -a -1 = 0
a² = a+1,a⁸ = (a+1)⁴
a¯¹ = a-1,a¯⁴ = (a-1)⁴
代数式a的8次方加｛7乘a的负4次方｝
= (a+1)⁴ + 7(a-1)⁴
= (a+1)⁴ -(a-1)⁴ + 8(a-1)⁴
= 8a(a+2)+8(a-1)⁴
=8(a² -1)(a+2)+8(a-1)⁴
= 8(a-1) [(a+1)(a+2) + (a-1)³]
= 8(a-1) (a³-2a²+6a+1)
= 8(a-1) (a - a²+6a+1)
= 8(a-1) 6a
=48(a²-a)
= 48
这道题肯定有许多种解法,我的解法是不是太麻烦了?