The tangent slope at P is 3x^2 and point P is at (x, x^3) right but how do i find an expression for Q. I tried using (x+h, (x+h)^3) but that doesnt work out in the end.

Let P be (h,h³).
Let Q be (k,k³).

The tangent line at P is y=mx+b, with m=3a² (as you already indicated). You can find 'b' in two ways:

Evaluate y at the point P and get an expression for b in terms of h.
Evaluate y at the point Q and get an expression for b in terms of h and k.

Since the b in both calculations is the same, you can equate the two expressions to get k in terms of h.

Try that, and if you still need help, post what you get and what you're stuck on.

The problem says that you construct a tangent (to y= x) at P and Q is the point where it again intersects the curve y= x³.

Writing P as (h,h³) and Q as (k,k³), the tangent line at P is y= (3h²)(x- h)+ h³. That will cross the curve at (k,k³) when k³= (3h²)(k-h)+ h³. From that k³- h³= (3h²)(k-h).

Of course, (k³- h³)/(k-h)= k²+kh+ h² this tells us that
k²+ kh+ h²= 3h²

Now determine the slope of the tangent line at Q (3k²) and compare it to 3h².

h³=(3h²)(h)+b
-2h³=b

For Point Q:

k³=3h²k+b
b=(k³)-(3h²k)

Indeed, since the whole point of a tangent line is that it passes through a certain point on a curve and has a certain slope there, it would be a really good idea to know the "point-slope" form of a line!

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so now we make them equal each other and this part i still dont understand. why do both b's equal each other? shouldnt they both be different because are we not talking about the tangent lines at P and Q? why are they both the same?

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Exactly. They are NOT the same. That's why I (1) found the equation of the tangent line at P. (2) found where that crosses the curve when x=Q (so I could get a relationship between P and Q) (3) found the slope of the tangent line at x= Q.