If $$ a^2-3a+1=0 $$ , find the value of the algebraic expression $$ 3a^3-8a^2+a+\dfrac{3}{a^2+1} $$

Question

Uniteasy Admin · Accepted Answer

$$ 3a^3-8a^2+a+\dfrac{3}{a^2+1} $$
$$ = 3a^3-9a^2+3a+a^2-2a+\dfrac{3}{a^2+1} $$
$$ =3a(a^2-3a+1) +a^2-2a+\dfrac{3}{a^2+1} $$
$$ = a^2-2a+\dfrac{3}{a^2+1} $$
$$ =a^2-3a+1+a-1+\dfrac{3}{a^2-3a+1+3a} $$
$$ =a-1+\dfrac{1}{a} $$
$$ =\dfrac{a^2-a+1}{a} $$
$$ =\dfrac{a^2-3a+1+2a}{a} $$
$$ =2 $$

Question

Answer