Polynomified

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

Now from the given quadratic we get $a^2=4a+3$ Thus, $\dfrac{4a+3+1}{a+1} = \dfrac{4(a+1)}{a+1} = 4$

The given expression can be simplified to $\dfrac{(a-1)(a^2+1)}{a^2-1}....................(1)$ .Now,we are also given that $a^2-4a-3=0\\ \Longrightarrow a^2-4a-4+1=0\\ \Longrightarrow a^2+1=4a+4\\ =4(a+1)$ .Substituting in $(1)$ gives $\dfrac{4(a+1)(a-1)}{a^2-1)}\\ =\dfrac{4(a^2-1)}{a^2-1}=4$ .

$a^{2}-4a-3 = 0$ has roots $a = 2 + \sqrt{7}$ and $a = 2 - \sqrt{7}$

With Factor Theorem we have $a-1$ as a factor of $a^{3}-a^{2}+a-1$ and it can be factorised into $(a-1)(a^{2}+1)$ .

Hence

$\frac{(a-1)(a^{2}+1)}{a^{2}-1}$ $=>$ $\frac{a^{2}+1}{a+1}$

Both values of $a$ give the value of the expression as $\boxed{4}$ and we are done .

separating the common things equalize the equation and then after solving we get 4 the answer

My solution was much messier... I found the roots as square root of two + 7, and plugged it in.

• the way without factorization
• if $a^2-4a-3$ then $a^2=4a+3$ put $4a+3$ whenever you see $a^2$
• $a^2-1=4a+3-1=4a+2$
• $\frac{a^3-a^2+a-1}{a^2-1}$ =
• $\frac{a.a^2-a^2+a-1}{a^2-1}$ = $\frac{a(4a+3)-(4a+3)+a-1}{4a+2}$
• = $\frac{4a^2+3a-4a-3+a-1}{4a+2}$ = $\frac{4(4a+3)+3a-4a-3+a-1}{4a+2}$
• = $\frac{16a+12+3a-4a-3+a-1}{4a+2}$ = $\frac{16a+8}{4a+2}$ = 4

Forgive any bad Latex coding I'm new to it but here's the solution....

Simplify both top and bottom of the equation giving us:

$\dfrac{(a-1)(a^2+1)}{(a-1)(a+1)}$

cancel out the $a-1$ which then gives us:

$\dfrac{a^2+1}{a+1}$

now using the information given in the question that $a^2-4a-3 = 0$ we look to the numerator...

$a^2+1=a^2-4a+4a+1+3-3$

$= a^2-4a-3+4a+4$

$=0+4a+4$

$=4(a+1)$

putting the numerator and denominator back together:

$\dfrac{4(a+1)}{(a+1)}$ = 4

a^2 - 4a - 3 = 0 implies that a^3 = 4a^2 + 3a.

So we substitute that into the expression to get

(3a^2 + 4a - 1)/(a^2 - 1)

It also implies that a^2 = 4a + 3 and 3a^2 = 12a + 9

Substituting both into the expression again, we get

(16a + 8)/(4a + 2) = 4.

[a (a^2 + 1) - (a^2 + 1)]/ (a^2 - 1) = [(a^2 + 1) (a - 1)]/ [(a + 1) (a - 1)]

= (a^2 + 1)/ (a + 1)

= 4 since (a^2 + 1) = 4 a + 4 = 4 (a + 1) as the quadratic equation given.