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Solution
$$a^3 + 3a^2 + 9a = 1$$
$$a(a^2 + 3a + 9)=1$$
$$a^2 + 3a + 9=1/a$$
$$(a^3-b^3)$$=$$(a-b)(a^2+ab+b^2)$$
for b=3
we have $$(a^3-3^3)$$=$$(a-3)(a^2+3a+9)$$
$$(a^3-27)$$=$$(a-3)(1/a)$$
$$a^3+(3/a)=1+27$$
$$a^3+(3/a)=28$$
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