Question

find the cube of the following binomial ;(i) (x+3y)3​

Answer 1

Answer:

(x^3+27y^3+9yx^2+27xy^2)

Step-by-step explanation:

=(x+3y)^3

=(x+3y)^2*(x+3y)^1

=(x^2+6xy+9y^2)*(x+3y)

=x*(x^2+6xy+9y^2)+3y*(x^2+6xy+9y^2)

=(x^3+6x^2y+9xy^2)+(3yx^2+18xy^2+27y^3)

=(x^3+6yx^2+3yx^2+9xy^2+18xy^2+27y^3)

=(x^3+27y^3+9yx^2+27xy^2)

I hope it helps, Mark me as Brainliest.