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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Updated on: 10-Oct-2022

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