If a function has a slant asymptote, it means that it is asymptotically equivalent to a certain line y=mx+q.
This means that
lim_{x\to\infty} f(x)-(mx+q)=0.
Divide by x:
lim_{x\to\infty} f(x)/x-(m+q/x)=0.
Since q/x\to 0, we have that
lim_{x\to\infty} f(x)/x-m=0.
And so m=lim_{x\to\infty} f(x)/x
Once m is known, we can find q by calculating
lim_{x\to\infty} f(x)-mx=q.
Now, let's do the calulations: in your case,
lim_{x\to\pm\infty} f(x)/x = 3, since we have
f(x)/x=\frac{3x^3-28x^2+54x-24}{x^3-8x^2+7x}
And the limits at \pm\infty are given by the ratio of the coefficients of the leading terms (i.e. x^3), which in this case are 3 and 1.
So, m=3. Now let's compute q:
\lim_{x\to\pm\infty} f(x)-mx = \frac{3x^3-28x^2+54x-24}{x^2-8x+7} - 3x
And since
\frac{3x^3-28x^2+54x-24}{x^2-8x+7} - 3x = \frac{3x^3-28x^2+54x-24-(3x^3-24x^2+21x)}{x^2-8x+7}
=\frac{-4x^2+33x-24}{x^2-8x+7}
And this function tends to -4 as x\to\pm\infty
