Question

How do you expand the binomial `(x-3y)^6` using the binomial theorem?

Answer 1

`(x-3y)^6=x^6-18x^5y+135x^4y^2-540x^3y^3+1215x^2y^4-1458xy^5+729y^6`.

Explanation:

The Binomial Theorem states that,

`(a+b)^n=""_nC_0a^(n-0)b^0+""_nC_1a^(n-1)b^1+""_nC_2a^(n-2)b^2+...`

`+""_nC_na^(n-n)b^n`.

We have, `a=x, b=-3y, n=6`. Hence,

`(x-3y)^6=""_6C_0x^(6-0)*(-3y)^0+""_6C_1x^(6-1)(-3y)^1+""_6C_2x^(6-2)(-3y)^2+""_6C_3x^(6-3)(-3y)^3+""_6C_4x^(6-4)(-3y)^4+""_6C_5x^(6-5)(-3y)^5+""_6C_6x^(6-6)(-3y)^6`.

Here, `""_6C_0=""_6C_6=1, ""_6C_1=""_6C_5=6`

`""_6C_2=""_6C_4=15, ""_6C_3=20`

`:. (x-3y)^6=x^6-18x^5y+135x^4y^2-540x^3y^3+1215x^2y^4-1458xy^5+729y^6`.