I have forgotten the formula to figure this out; can you solve it? (x+1)^3?

Question

I need help figuring out how to solve this. (x+1)^3
I know the answer is not (x+1)*(x+1)*(x+1)... There's a special formula to solve this, though I've forgotten it.

Thanks!

Answer 1

You can use the long way of expanding out each term and adding:
(x+1)(x+1) = x² + 2x + 1
(x+1)(x+1)(x+1) = (x + 1)(x² + 2x + 1)
= x(x² + 2x + 1) + 1(x² + 2x + 1)
= x^3 + 2x² + x + x² + 2x + 1
= x^3 + 3x² + 3x + 1

But there is a much easier method to solve this called "binomial expansion". First you need to know Pascal's triangle. For (x+1)^3 you need row 3 (the first row is row 0).

Pascal's triangle:
Row 0: ..... 1
Row 1: .... 1 1
Row 2: ....1 2 1
Row 3: ...1 3 3 1 <---
Row 4: ..1 4 6 4 1
Row 5: 1 5 10 10 5 1

Now you need to make all the combinations of x and 1 in descending order. First you need x^3, then x^2, then x, then 1.

Now put each number (1, 3, 3, 1) from that row of the Pascal's triangle in front of each term.

So (x + 1)^3 = x^3 + 3x² + 3x + 1

You can use it to do even higher powers easily... for example:
(x + 1)^7 = x^7 + 7x^6 + 21x^5 + 35x^4 + 35x^3 + 21x² + 7x + 1

I think that is the " special formula" you were looking for. With it you can quickly see:
(x + 1)^3 = x^3 + 3x² + 3x + 1

Answer 2

x^3 + 1^3 = x^3 +1

Answer 3

i know what you mean. the formula goes something like

(a+b)^c = [a^(c) x (b^0)] + [a^(c-1) x (b^1)] + [a^(c-2) x b^2] .... [a^(c-c) x b^c]

basically the first set is the [a^c x b^0] and then the rest has c decreasing by 1 until it reaches 0, and b's interger increasing from 0 until it reaches c.


your problem/answer is:
(x+1)^3 = x^3 + 3x^2 + 3x + 1

you know exponent is 3, so to get there the answer goes like

[(x^3)(1^0)] + [(x^(3-1))(1^1)] + [(x^(3-2))(1^2)] + [(x^(3-3))(1^3)]

i hope this helps!


oh yea and if the original problem is subtraction you just alternate the signs in the answer

Answer 4

x^3+x^2+12

Answer 5

(x+1)(x+1)(x+1)
do the first two parenthesis first then multiple each by the second parenthesis
(x2 + 2x +1)(x + 1)
x3 + 2x2 + x + x2 + 2x + 1
combine like terms
x3 + 3x2 + 3x + 1
note: there isn't a way to use exponents here so the 2 after the x and the 3 afer the 3x are exponents meaning cubed and squared

Answer 6

Work it out. (x+1)(x+1)(x+1) = (x^2 + 2x + 1)(x+1) = x^3 + 3x^2 + 3x + 1

Answer 7

I know what you are asking. The general formula for (X+1)^n=
X^n+n*X^(n-1)+n*X^(n-2)+
n*X^(n-3)+................+n*X^(1) + 1.
Hope this is what you required. Else get back again

Answer 8

x^3+3x^2+3x+1

Answer 9

its called foil. You times the first set by the rest of it and just follow the line so inner times inner outer times outer.