How do you expand ${(3 x + 3 y)}^{3}$ $(3x+3y)^3$ ?

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Answer 1 · 2016-08-14T17:24:47

$27 x^{3} + 81 x^{2} y + 81 x y^{2} + 27 y^{3}$ $27x^3+81x^2y+81xy^2+27y^3$ .

We know that $: {(a + b)}^{3} = a^{3} + b^{3} + 3 a b (a + b)$ $: (a+b)^3=a^3+b^3+3ab(a+b)$

Now, ${(3 x + 3 y)}^{3} = {3 (x + y)}^{3} = 3^{3} \cdot {(x + y)}^{3}$ $(3x+3y)^3={3(x+y)}^3=3^3*(x+y)^3$

$= 27 {x^{3} + y^{3} + 3 x y (x + y)}$ $=27{x^3+y^3+3xy(x+y)}$

$= 27 (x^{3} + y^{3} + 3 x^{2} y + 3 x y^{2})$ $=27(x^3+y^3+3x^2y+3xy^2)$

$= 27 x^{3} + 81 x^{2} y + 81 x y^{2} + 27 y^{3}$ $=27x^3+81x^2y+81xy^2+27y^3$ .

How do you expand (3x+3y)3(3x+3y)^3?