How do you expand the binomial ${(2 x - y)}^{6}$ $(2x-y)^6$ using the binomial theorem?

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How do you expand the binomial ${(2 x - y)}^{6}$ $(2x-y)^6$ using the binomial theorem?

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Answer 1 · 2016-08-29T15:36:02

${(2 x - y)}^{6} = 64 x^{6} - 192 x^{5} y + 240 x^{4} y^{2} - 160 x^{3} y^{3} + 60 x^{2} y^{4} - 12 x y^{5} + y^{6}$ $(2x - y)^6 = 64x^6 - 192x^5y + 240x^4y^2 - 160x^3y^3 + 60x^2y^4 - 12xy^5 + y^6$

Explanation:

${(2 x - y)}^{6}$ $(2x - y)^6$

The binomial theorem states that for any binomial ${(a + b)}^{n}$ $(a + b)^n$ , the general expansion is given by ${(a + b)}^{n} = {t w o}_{n} C_{r} \times a^{n - r} \times b^{r}$ $(a + b)^n = color(white)(two)_nC_r xx a^(n - r) xx b^r$ , where $r$ $r$ is in ascending powers from $0$ $0$ to $n$ $n$ and $n$ $n$ is in descending powers from $n$ $n$ to $0$ $0$ .

$= color(white)(two)_6C_0(2x)^6(-y)^0 + color(white)(two)_6C_1(2x)^5(-y)^1 + color(white)(two)_6C_2(2x)^4(-y)^2 + color(white)(two)_6C_3(2x)^3(-y)^3 + color(white)(two)_6C_4(2x)^2(-y)^4 + color(white)(two)_6C_5(2x)^1(-y)^5 + color(white)(two)_6C_6(2x)^0(-y)^6$

$=64x^6 - 192x^5y + 240x^4y^2 - 160x^3y^3 + 60x^2y^4 - 12xy^5 + y^6$

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How do you expand the binomial ${(2 x - y)}^{6}$ $(2x-y)^6$ using the binomial theorem?

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