（x-1/2）的平方*（x的平方+1/4）的平方*（x+1/2）的平方？？求解？？、
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(x-1/2)²(x²+1/4)²(x+1/2)²
=(x²-x+1/4)(x²+1/4)²(x²+x+1/4)
=[(x²+1/4-x)(x²+1/4+x)](x²+1/4)²
=[(x²+1/4)²-x²](x²+1/4)²
=[x^4+1/2·x²+1/16-x²]²(x²+1/4)²
=(x^4-1/2·x²+1/16)²(x²+1/4)²
=(x²-1/4)²(x²+1/4)²
=[(x²-1/4)(x²+1/4)]²
=(x^4-1/16)²
=x^8-1/8·x^4+1/256
=(x²-x+1/4)(x²+1/4)²(x²+x+1/4)
=[(x²+1/4-x)(x²+1/4+x)](x²+1/4)²
=[(x²+1/4)²-x²](x²+1/4)²
=[x^4+1/2·x²+1/16-x²]²(x²+1/4)²
=(x^4-1/2·x²+1/16)²(x²+1/4)²
=(x²-1/4)²(x²+1/4)²
=[(x²-1/4)(x²+1/4)]²
=(x^4-1/16)²
=x^8-1/8·x^4+1/256
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