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已知:a2-b2=(a-b)(a+b);a3-b3=(a-b)(a2+ab+b2);a4-b4=(a-b)(a3+a2b+ab2+b3);a5-b5=(a-b)(a4+a3b+a2b2+ab3+b4)按此规律,则:(1)a6-b6=(a-b);(2)若a−1a=3,请你根据上述规律求出

题目详情
已知:a2-b2=(a-b)(a+b);a3-b3=(a-b)(a2+ab+b2);a4-b4=(a-b)(a3+a2b+ab2+b3);a5-b5=(a-b)(a4+a3b+a2b2+ab3+b4)按此规律,则:
(1)a6-b6=(a-b)______;
(2)若a−
1
a
=3,请你根据上述规律求出代数式a3−
1
a3
的值.
▼优质解答
答案和解析
(1)根据规律可知,a6-b6=(a-b)(a5+a4b+a3b2+a2b3+ab4+b5);

(2)a3−
1
a3

=(a-
1
a
)(a2+a•
1
a
+
1
a2

=(a-
1
a
)(a2+a•
1
a
+
1
a2

=(a-
1
a
)(a2+
1
a2
+1)
=(a-
1
a
)(a2+
1
a2
+2a•
1
a
-2a•
1
a
+1)
=(a-
1
a
)[(a2+
1
a2
-2a•
1
a
)+2+1]
=(a-
1
a
)[(a-
1
a
2+3]
=3×(32+3)
=3×12
=36.