Question

Simplify $a\left(a^2+a+1\right)+5$ and find its value for (i) $a$=$0$, (ii) $a$=$1$
(iii) $a$ = $-1$

Answer 1

Given expression is

$a(a^2+a+1)+5$ = $a\times a^2+a\times a+a\times 1+5$

$=a^3+a^2+a+5$

(i) For $a=0,a^3+a^2+a+5=(0{)}^3+(0{)}^2+\left(a\right)+5$

$=0+0+0+5=5$

(ii) For $a=1,a^3+a^2+a+5=(1{)}^3+(1{)}^2+\left(1\right)+5$

$=1+1+1+5=8$

(iii) For $a=-1,a^3+a^2+a+5=(-1{)}^3+(-1{)}^2+\left(1\right)+5$

$=-1+1-1+5=4$