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Question

1+a1111+aa111+a=a3+3a2

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Solution

=1+a1111+a1111+a=1+a 1+a111+a - 11111+a +111+a11 Expanding=(1+a)(1+a)2 -1-1(1 + a - 1) + (1 - 1 - a)=(1+a)[1 + a2 + 2a - 1] -a -a=1 + a + a2 + a3 + 2a + 2a2 - 2a= a3 + 3a2

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