Question

expand ..... the following​

Answer 1

Answer:

Use the binomial expansion theorem to find each term. The binomial theorem states (a+b)n=n∑k=0nCk⋅(an−kbk)(a+b)n=∑k=0n⁡nCk⋅(an-kbk).

3∑k=03!(3−k)!k!⋅(2x)3−k⋅(−2)k∑k=03⁡3!(3-k)!k!⋅(2x)3-k⋅(-2)k

Expand the summation.

3!(3−0)!0!⋅(2x)3−0⋅(−2)0+3!(3−1)!1!⋅(2x)3−1⋅(−2)+3!(3−2)!2!⋅(2x)3−2⋅(−2)2+3!(3−3)!3!⋅(2x)3−3⋅(−2)33!(3-0)!0!⋅(2x)3-0⋅(-2)0+3!(3-1)!1!⋅(2x)3-1⋅(-2)+3!(3-2)!2!⋅(2x)3-2⋅(-2)2+3!(3-3)!3!⋅(2x)3-3⋅(-2)3

Simplify the exponents for each term of the expansion.

1⋅(2x)3⋅(−2)0+3⋅(2x)2⋅(−2)+3⋅(2x)⋅(−2)2+1⋅(2x)0⋅(−2)31⋅(2x)3⋅(-2)0+3⋅(2x)2⋅(-2)+3⋅(2x)⋅(-2)2+1⋅(2x)0⋅(-2)3

Simplify each

Answer 2

Answer:

8(x⁶-1/x³+3-3x²/x)

Explanation:

Expand using the identity (a-b)³= a³-b³-3a²b+3ab²