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

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Question

(A) Given expression: $p(p^{2} + 3p + 3) = p^{3} + 3p^{2} + 3p$.To simplify this,we can add and subtract $1$ to complete the cube formula: $(p+1)^{3}

Vedclass · Accepted Answer

$999999$

Vedclass · Answer

(A) Given expression: $p(p^{2} + 3p + 3) = p^{3} + 3p^{2} + 3p$.To simplify this,we can add and subtract $1$ to complete the cube formula: $(p+1)^{3} = p^{3} + 3p^{2} + 3p + 1$.Therefore,$p^{3} + 3p^{2} + 3p = (p+1)^{3} - 1$.Substitute $p = 99$ into the expression:$(99 + 1)^{3} - 1 = 100^{3} - 1$.$100^{3} = 1000000$.$1000000 - 1 = 999999$.

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

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