If p=99, then the value of p(p^{2}+3p+3) is | vedclass

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(C) Given that $p=99.$We need to find the value of $p(p^{2}+3p+3).$Expanding the expression,we get $p^{3}+3p^{2}+3p.$To simplify this,we add and subtr

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$999999$

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(C) Given that $p=99.$We need to find the value of $p(p^{2}+3p+3).$Expanding the expression,we get $p^{3}+3p^{2}+3p.$To simplify this,we add and subtract $1$ to complete the cube formula $(a+b)^{3} = a^{3}+3a^{2}b+3ab^{2}+b^{3}.$So,$p^{3}+3p^{2}+3p+1-1 = (p+1)^{3}-1.$Substituting $p=99,$ we get $(99+1)^{3}-1 = (100)^{3}-1.$$(100)^{3} = 1000000.$Therefore,$1000000-1 = 999999.$

If $p=99,$ then the value of $p(p^{2}+3p+3)$ is

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