How do I expand (x+3)^3?

Question

Show all steps please.

Answer 1

(x+3)(x+3)(x+3)
(x^2+3x+3x+9)(x+3)
thn multiply (x+3) into the polynomial one at a time

Answer 2

(a + b)^3 = a^3 + 3 ab (a +b) + b^3
Put a = x and b = 3
x^3 + 3 × x × 3 (x + 3) + 3^3
= x^3 + 9 x^2 + 27 x + 27

Answer 3

All the ^3 means is just to take that number and multiply it three times so itll just be like sayin 3^3 is 3x3x3=27 but you dont want to solve it all you want to do is expand it so itll be

(x+3) (x+3) (x+3)

Answer 4

Binomial theorem says (x+y)^3 =

x^3 + 3(x^2)(y) + 3(x)(y^2) + y^3.

Let y = 3:

x^3 + 3(x^2)(3) + 3(x)(3^2) + 3^3.

x^3 + 9(x^2) + 27x + 27.

Answer 5

(x+3) times by 3
3(x+3) =
3x + 9

Answer 6

multiply (x+3)(x+3)(x+3)
(x^2+6x+9) (using foil)
then multiply the polynomial by (x+3)
x^3+6x^2+9x+3x^2+18x+27 (then combine like terms)
x^3+9x^2+27x+27

Answer 7

Just do this:
(x+3)(x+3)
(x+3)=
(x^2+6x+9)
(x+3)=
x^3+6x^2+9x
+3x^2+18x+27=
x^3+9x^2+27x+27

Answer 8

(x+3)^3 = (x+3)*(x+3)*(x+3)
= (x^2+3x+3x+9)*(x+3)
= (x^2+6x+9)*(x+3)
= x^3+6x^2+9x+3x^2+18x+27
= x^3+9x^2+27x+27

Answer 9

(x+3)^3
(x+3) (x+3)^2
(x+3) (x+3) (x+3)
x^2+3x+3x+9 (x+3)
x^2+6x+9 (x+3)
(x^3+6x^2+9x)+(3x^2+18x+27)
x^3+9x^2+27x+27

Answer 10

make 3 columns. 1st column is row 3 of Pascal's triangle. 2nd column is descending powers of x. 3rd column is ascending powers of 3:

1 ... x³ ..... 1
3 ... x² ..... 3
3 ... x ...... 9
1 ... 1 ..... 27

multiply each row together and add terms:
x³ + 9x² + 27x + 27