Solve for (x+y)^3

Question

Please show me work of how to solve this problem. I've been beating myself up! Thanks.

Answer 1

(x+y)² =(x+y)(x+y)    Then you FOIL (First, outer, inner, last)

(x+y)² =(x+y)(x+y) = xx + xy + xy + yy [and when you combine like terms] = x² + 2xy + y²

(x+y)³ = (x² + 2xy + y²)(x+y) Then you FOIL (First, outer, inner, last)

(x+y)3 = (x² + 2xy + y²)(x+y) = x²x +2xxy + xy² + x²y + 2xyy + y²y [and when you combine like terms] = x³ + 3x²y+ 3xy² + y³

Answer 2

To make the exponents, use the shortcut for superscripts: Press CTRL+SHIFT+=; then release the 3 keys and type whatever exponent you wish to insert.

Answer 3

(x+y)³ expanded has 4 terms, 1 more than the exponent, x³ x²y xy² and y³ x is decreasing from 3 to 0 from left to right, as y increases from 0 to 3. Any number or variable to the 0 power is 1.

Then you need the coefficients for each of the 4 terms. You can read that off Pascal's triangle:

1

11

121

1331

1, 3, 3 and 1 are the coefficients that go with each of the 4 terms to get x^3 +3x^2y + 3xy^2 + y^3

Or if you know combinations from probability theory, the coefficients are (3/3), (3/2), (3/1) and (3/0)

(n/r) = n!/r!(n-r!) (3/3) = 3!/3!0! = 1, (3/2)= 3!/2!1! = 3, (3/1) = 3!/1!2! = 3 and (3/0) = 3!/0!3! = 1

0! = 1, 1! = 1, 2! = 2x1 = 2, 3! =3x2x1 = 6

x³ + 3x²y + 3xy² + y³

Each number in Pascal's triangle is the sum of the 2 digits just above it, to the right & left.

Or you could just multiply the 3 factors together (x+y)(x+y)(x+y) =(x^2+2y+y^2)(x+y)= (x^3+3x^2y+3xy^2+y^3)

Or you might just remember the formula: (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 and replace a with x and b with y

People remember the formula (a+b)^2 = a^2 +2ab + b^2, for the cubic it's similar

power of a decreases by 1 each term, 2, 1, then an implicit 0 while b increases by 1 from an implicit 0 to 1, to 2

for the cubic expression powers of a decrease from 3, to 2, to 1, to 0

while b increases from 0, to 1, to 2, to 3

Answer 4

If you want to do the caclulations alot quicker, use Pascal's triangle. Let's say you have (x+y)^n. If you take n! and divide it by (n-r)!*r! then you will have the coefficient to each part of the equation. ! is factorial. Lets say you have 5! = 5*4*3*2*1. r would equal the position of the coefficient. In the case of (x+y)^3 the numbers on pascals triangle are 1 3 3 1. Which means the answer is 1*x^3 + 3*(x^2)(y) + 3*(x)(y^2) + 1*y^3. Do you see a pattern with x and y. The numbers in pascals triangle represents how many different combinations can be taken from a limited number of something. (Ex. How many ways can you take 3 employees from 5 employees). The answer would be 10 ways if you use the equation. Here is a more detailed explanation of Factorial. Imagine you have five pieces of candy. If you choose one then you can only choose four more. If you choose another piece of candy then you can only choose three more pieces and so on and so on until you arrive at 1. (5*4*3*2*1). This method will save you so much time if you have to deal with large exponents. If you don't understand I will be happy to explain with more clarity.