Expand (1 + 𝑥 5 )^3 in ascending powers of 𝑥, up to and including the term in 3 x .

i know how to expand the equation but im not sure what the question is asking me to do

YEEEEEET Nov 10, 2018

#3**+1 **

Expand the following:

(x^5 + 1)^3

Expand (x^5 + 1)^3 using the binomial expansion theorem.

(x^5 + 1)^3 | = | sum_(k=0)^3 binomial(3, k) (x^5)^(3 - k) 1^k |

| = | binomial(3, 0) (x^5)^3 1^0 + binomial(3, 1) (x^5)^2 1^1 + binomial(3, 2) (x^5)^1 1^2 + binomial(3, 3) (x^5)^0 1^3 | : invisible comma

binomial(3, 0) x^15 + binomial(3, 1) x^10 + binomial(3, 2) x^5 + binomial(3, 3)

Evaluate the binomial coefficients by looking at Pascal's triangle.

The binomial coeffients comprise the 4^th row of Pascal's triangle:

1 | | 3 | | 3 | | 1

(x^5)^3 + 3 (x^5)^2 + 3 x^5 + 1

For all positive integer exponents (a^n)^m = a^(m n). Apply this to (x^5)^2.

Multiply exponents. (x^5)^2 = x^(5×2):

(x^5)^3 + 3 x^(5×2) + 3 x^5 + 1

Multiply 5 and 2 together.

5×2 = 10:

(x^5)^3 + 3 x^10 + 3 x^5 + 1

For all positive integer exponents (a^n)^m = a^(m n). Apply this to (x^5)^3.

Multiply exponents. (x^5)^3 = x^(5×3):

x^(5×3) + 3 x^10 + 3 x^5 + 1Multiply 5 and 3 together.

**x^15 + 3x^10 + 3x^5 + 1**

Guest Nov 10, 2018