The expression $x^3-4x^2+ax+b$, where $a$ and $b$ are constants, has a factor of $x+1$ and leaves a remainder of $-60$ when divided by $x+3$.

1. Apply the Factor Theorem to form the first equation.

Let $P(x) = x^3 - 4x^2 + ax + b$.
Since $x+1$ is a factor, $P(-1) = 0$.
$(-1)^3 - 4(-1)^2 + a(-1) + b = 0$
$-1 - 4 - a + b = 0$
$-a + b = 5 \quad \text{--- (1)}$

2. Apply the Remainder Theorem to form the second equation.

Since the remainder is $-60$ when divided by $x+3$, $P(-3) = -60$.
$(-3)^3 - 4(-3)^2 + a(-3) + b = -60$
$-27 - 36 - 3a + b = -60$
$-63 - 3a + b = -60$
$-3a + b = 3 \quad \text{--- (2)}$

3. Solve the simultaneous equations to find $a$ and $b$.

Subtracting (2) from (1):
$(-a + b) - (-3a + b) = 5 - 3$
$2a = 2$
$a = 1$

Substitute $a = 1$ into (1):
$-1 + b = 5$
$b = 6$

1. Find the values of $a$ and $b$ from the conditions in part (a)

Let $P(x) = x^3 - 4x^2 + ax + b$.
Since $x+1$ is a factor, $P(-1) = 0$:
$(-1)^3 - 4(-1)^2 + a(-1) + b = 0 \implies -a + b = 5$
Since division by $x+3$ leaves a remainder of $-60$, $P(-3) = -60$:
$(-3)^3 - 4(-3)^2 + a(-3) + b = -60 \implies -3a + b = 3$
Solving simultaneously:
$(-a + b) - (-3a + b) = 5 - 3 \implies 2a = 2 \implies a = 1$
$b = 5 + 1 = 6$

2. Substitute $a$ and $b$ into the equation and factorise out the known linear factor

The equation is $x^3 - 4x^2 + x + 6 = 0$.
Since $x+1$ is a factor, we can write:
$x^3 - 4x^2 + x + 6 = (x+1)(x^2 + cx + d)$
By comparing coefficients:
Constant term: $1 \times d = 6 \implies d = 6$
Coefficient of $x^2$: $1 + c = -4 \implies c = -5$
So, $x^3 - 4x^2 + x + 6 = (x+1)(x^2 - 5x + 6) = 0$

3. Factorise the quadratic factor

$(x+1)(x-2)(x-3) = 0$

4. Solve for $x$

$x = -1$, $x = 2$, or $x = 3$

1. Rewrite the given exponential equation to resemble a polynomial

The given equation is $9^x + 1 = 4(3^x) - 6(3^{-x})$.
Multiply throughout by $3^x$:
$9^x(3^x) + 3^x = 4(3^x)(3^x) - 6$
$(3^2)^x(3^x) + 3^x = 4(3^{2x}) - 6$
$3^{3x} + 3^x = 4(3^{2x}) - 6$
$(3^x)^3 - 4(3^x)^2 + (3^x) + 6 = 0$

2. Relate the rearranged equation to the polynomial from part (b)

From part (a), $a = 1$ and $b = 6$. The equation in part (b) is $x^3 - 4x^2 + x + 6 = 0$.
Comparing $(3^x)^3 - 4(3^x)^2 + (3^x) + 6 = 0$ with $x^3 - 4x^2 + x + 6 = 0$, we can see that substituting $x$ with $3^x$ in the polynomial equation gives the exponential equation.
Therefore, the solutions for $3^x$ are the roots found in part (b). We can solve for $x$ by taking logarithms: $x = \log_3(\text{root})$.