How do I find (2-h)^3?

Question

Could you please help me figure out how to do this step by step?

Answer 1

(2-h)^3 there's a couple of ways you can do this but I think the easiest is to remember that this also equaly
(2 - h) (2 - h) (2 - h)
= (2 - h)^2 (2-h)
now expand the first bracket the same as any quadratic
= (4 - 4h + h^2) (2 - h)
now multiply these two together, everything in the first bracket times everything in the second bracket, so that
= 8 - 4h - 8h + 4h^2 + 2h^2 - h^3
now group all the like terms together
= 8 - 12h + 6h^2 - h^3

Answer 2

I suppose you know that "to the power 3" means multiply out so there a total of 3 identical factors.

You didn't say if you know the value of "h".
If you do, then you know the value of 2-h.
Multiplying it by itself twice will give you some number.
That's the answer for a known value of "h".

I suspect you don't know the value of "h".
Multiplying 2-h by itself should be easy:

2-h
*(2-h) ..... * means multiply
-------- start multiplying from the right, just like a 2digit no.
.............. even tho these are nos. of unknown nos. of digits.
-2h+h^2 is the product of -h and the top expression
4-2h is the product of 2 and the top expression
----------- adding them should be simple:
4-4h+h^2 is the first intermediate answer
*(2-h) ..... has to be done again
-------- again, multiply from the right
-4h+4h^2-h^3 is the product of -h and the top expression
8-8h+2h^2 is the product of 2 and the top expression
------------- we add like terms...
8-12h+6h^2-h^3 is the result.

Answer 3

mathassignm was almost right but missed a sign. It should be

(2-h)(h^2-4h+4)
2h^2-8h+8-h^3 + 4h^2-4h

8 - 12h + 6h^2 - h^3

Answer 4

Set it up as multiplication, or (2-h)*(2-h)*(2-h) since it is to the third power. Then use the FOIL (First, Outside, Inside, Last) process on the first two sets in parentheses, or (2-h)*(2-h). Your answer will come out to (4 - 4h + h^2). You then just take that answer times (2-h), and you will have your final answer, which comes out to be 8 - 12x + 6x^2 - x^3.

Answer 5

(2-h)^3

Expand to binomials without exponents:
(2-h)(2-h)(2-h)

Multiply two of these using FOIL:
(4 - 2h - 2h + h²)(2 - h) = (4 - 4h + h²)(2 - h)

Now, use distributive property and multiply first polynomial by both terms in the second one:
(4-4h+h²)2 - (4-4h+h²)h
8 - 8h +2h² - 4h +16h² - h^3

Combine like terms and sort from highest degree to lowest:
-h^3 + 18h² - 12h + 8

Answer 6

You put together powers of 2 from 3 down to 0, powers of -h from 0 up to 3, and row 3 of Pascal's Triangle (1,3,3,1):

1(2^3)(-h)^0 + 3(2^2)(-h)^1 + 3(2^1)(-h)^2 + 1(2^0)(-h)^3 =
8 - 12h + 6h² - h^3

Answer 7

First multiply
(2-h)(2-h)
4-4h+h^2, then multiply again by 2-h
(2-h)(h^2-4h+4)
2h^2-8h+8-h^3-4h^2-4h
combine
-h^3-2h^2-12h+8

Answer 8

(2 - h)^3

= (2 - h) (2 - h)^2

= (2 - h) (4 - 4h + h^2)

= 2 (4 - 4h + h^2) - h (4 - 4h + h^2)

= 8 - 8h + 2h^2 - 4h + 4h^2 - h^3

= 8 -12h + 6h^2 - h^3

Answer 9

(2 - h) (2 - h) (2 - h)
(2 - h) (4 - 4h + h²)
8 - 8h + 2h² - 4h + 4h² - h³
8 - 12h + 6h² - h³

Answer 10

(2-h)(2-h)(2-h)=
(4-4h+h^2)(2-h)=
(8-8h+2h^2)-4h+4h^2-h^3=
8-12h+6h^2-h^3