(x+y) to the 5th power

look at row 5 of Pascal's triangle

1 5 10 10 5 1

(x+y)^5
= x^5 + 5x^4 y + 10x^3 y^2 + 10x^2 y^3 + 5xy^4 + y^5

To calculate (x+y) raised to the 5th power, you can use the binomial theorem or the distributive property. Here's the step-by-step explanation of finding the answer:

Step 1: Write out the expression:
(x+y)⁵

Step 2: Expand the expression using the binomial theorem or the distributive property. In this case, let's use the distributive property.
(x+y)⁵ = (x+y)(x+y)(x+y)(x+y)(x+y)

Step 3: Multiply each term of the expression using the distributive property. Start with the first term of the expanded expression, which is (x+y)(x+y)(x+y)(x+y):

(x+y)(x+y)(x+y)(x+y) =
(x+y)(x^2 + 2xy + y^2)(x+y)(x+y) =
(x^3 + 2x²y + xy² + x²y + 2xy² + y³)(x+y)(x+y) =
(x^3 + 3x²y + 3xy² + y³)(x+y)(x+y)

Step 4: Continue multiplying each term using the distributive property:
(x^3 + 3x²y + 3xy² + y³)(x+y)(x+y) =
(x^4 + 3x³y + 3x²y² + xy³ + x³y + 3x²y² + 3xy³ + y⁴)(x+y) =
(x^4 + 4x³y + 6x²y² + 4xy³ + y⁴)(x+y)

Step 5: Multiply the last term:
(x^4 + 4x³y + 6x²y² + 4xy³ + y⁴)(x+y) =
x^5 + 4x⁴y + 6x³y² + 4x²y³ + xy⁴ + x⁴y + 4x³y² + 6x²y³ + xy⁴ + y⁵

Step 6: Simplify the terms:
x^5 + x⁴y + 4x⁴y + 4x³y² + 6x³y² + 6x²y³ + 6x²y³ + 4xy⁴ + xy⁴ + y⁵

Step 7: Combine similar terms:
x^5 + 5x⁴y + 10x³y² + 10x²y³ + 5xy⁴ + y⁵

Therefore, (x+y)⁵ is equal to x^5 + 5x⁴y + 10x³y² + 10x²y³ + 5xy⁴ + y⁵.