Help with algebra? Simplifying a function, kind of?

Question

* = raised to the power of
Ok sooo it starts with (x+h)*3 - 2 - (x*3 - 2)

Then you cube the (x+h), right, and that ends up being

hx*3 + 2h*2x*2 = h*3x - 2 - x*3 + 2

So the twos cancel out and it's just

hx*3 + 2h*2x*2 = h*3x - x*3

right?

Oh yeah and this has all been over h

So the book has the answer as 3x*2 + 3xh + h*2
But I have no idea how...

Help?

Answer 1

(x + h)³ - 2
(x + h)(x² + 2hx + h²) - 2
x³ + 3hx² + 3h²x + h³ - 2 = A say
B = x³ - 2
A - B = 3hx² + 3h²x + h³
(A - B) / h = 3x² + 3hx + h²
(A - B) / h = 3x² as h--> 0

Answer 2

i tried to expand (x+h)*3, the result is :
x*3 + 3hx*2 + 3xh*2 + h*3

then combine it withe the function:
x*3 + 3hx*2 + 3xh*2 + h*3 - 2 - (x*3 - 2)

so twos and x*3 is eliminated,
3hx*2 + 3xh*2 + h*3
= h(3x*2 + 3xh*2 + h*3)

if this all over h, then h is eliminated, right...
so the book's answer is make sense...
the final result is 3x*2 + 3xh*2 + h*3

Answer 3

You didn't cube (x+h)^3 correctly. It sould be
x^3 + 3hx^2 + 3xh^2 + h^3

So (x+h)^3 - 2 - (x^3 - 2) =
(x+h)^3 - 2 - x^3 + 2 =
(x+h)^3 - x^3 =
3hx^2 + 3xh^2 + h^3

Answer 4

(x+h)^3 = (x+h)(x+h)^2 = (x+h)(x^2 + 2hx + h^2) = x^3 + 2hx^2 + xh^2 + hx^2 + 2xh^2 + h^3 = x^3 + 3hx^2 + 3xh^2 + h^3

so

(x+h)^3 - 2 - (x^3 - 2)
(x+h)^3 - x^3
3hx^2 + 3xh^2 + h^3
h (3x^2 + 3xh +h^2)

Answer 5

this question is really tricky? hmmmmm
i also dont understand how the book got that answer?
i got hx*2+2hx*2+2h*2x+h*2x+h*3

Answer 6

(x+h)³ = x³ + 3hx² + 3h²x + h³

So the entire thing equals:
= x³ + 3hx² + 3h²x + h³ - x³ + 2 - 2
= 3hx² + 3h²x + h³
= h(3x² + 3hx + h²)