Question

expand (6x+y/3)^ please help..​

Answer 1

Answer:

a+b)n=n∑k=0nCk⋅(an−kbk)(a+b)n=∑k=0n⁡nCk⋅(an-kbk).

3∑k=03!(3−k)!k!⋅(6x)3−k⋅(y)k∑k=03⁡3!(3-k)!k!⋅(6x)3-k⋅(y)k

Expand the summation.

3!(3−0)!0!⋅(6x)3−0⋅(y)0+3!(3−1)!1!⋅(6x)3−1⋅(y)+3!(3−2)!2!⋅(6x)3−2⋅(y)2+3!(3−3)!3!⋅(6x)3−3⋅(y)33!(3-0)!0!⋅(6x)3-0⋅(y)0+3!(3-1)!1!⋅(6x)3-1⋅(y)+3!(3-2)!2!⋅(6x)3-2⋅(y)2+3!(3-3)!3!⋅(6x)3-3⋅(y)3

Simplify the exponents for each term of the expansion.

1⋅(6x)3⋅(y)0+3⋅(6x)2⋅(y)+3⋅(6x)⋅(y)2+1⋅(6x)0⋅(y)31⋅(6x)3⋅(y)0+3⋅(6x)2⋅(y)+3⋅(6x)⋅(y)2+1⋅(6x)0⋅(y)3

Simplify each term.