Simplify $a(a^{2}+a+1)+5$ and find its value for (i) $a=0$, (ii)$a=1$.
Given :
The given expression is $a(a^{2}+a+1)+5$.
To do :
We have to find the values for (i) $a=0$, (ii)$a=1$.
Solution :
(i) $a=0$ at $a(a^{2}+a+1)+5$,
$=0(0^2+0+1)+5$
$=0+5=5$
Therefore, the value of $a(a^{2}+a+1)+5$ for $a=0$ is 5.
(ii)$a=1$ at $a(a^{2}+a+1)+5$
$= 1 (1^2+1+1)+5$
$=1(1+1+1)+5$
$=1(3)+5$
$=3+5=8$
Therefore, the value of $a(a^{2}+a+1)+5$ for $a=1$ is 8.
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