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Solution
Given : $$p$$ = 99
To find : $$p(p^{2} + 3p + 3)$$
= $$p^3 + 3p^2 + 3p$$
= $$(p^3 + 3p^2 + 3p + 1) - 1$$
= $$(p + 1)^3 - 1$$
= $$100^3 - 1$$
= 1000000 - 1 = 999999
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