How do you expand the binomial $(2x-y^3)^7$ using the binomial theorem?

Question

3712 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-30T18:07:34

$128x^7-448x^6y^3+672x^5y^6-560x^4y^9+280x^3y^12-84x^2y^15$
$+14xy^18-y^21$

For the expansion of $(x+y)^n$ we have:

$sum_(r=0)^n((n),(r))x^(n-r)y^r$

Where:

$((n),(r))=color(white)(0)^nC_(r)=(n!)/((r!(n-r)!)$

$(2x-y^3)^7$

$((7),(0))(2x)^7(-y^3)^0+((7),(1))(2x)^6(-y^3)^1+((7),(2))(2x)^5(-y^3)^2$

$+((7),(3))(2x)^4(-y^3)^3+((7),(4))(2x)^3(-y^3)^4+((7),(5))(2x)^2(-y^3)^5$

$+((7),(6))(2x)^1(-y^3)^6+((7),(7))(2x)^0(-y^3)^7$

Using $\ \ \ (n!)/((r!(n-r)!)$

$128x^7-448x^6y^3+672x^5y^6-560x^4y^9+280x^3y^12-84x^2y^15$
$+14xy^18-y^21$

How do you expand the binomial (2x-y^3)^7(2x-y^3)^7 using the binomial theorem?