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

Answer 1

Ariana Alvarez

#27x^3+81x^2y+81xy^2+27y^3#.

We know that # : (a+b)^3=a^3+b^3+3ab(a+b)#

Now, #(3x+3y)^3={3(x+y)}^3=3^3*(x+y)^3#

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

#=27(x^3+y^3+3x^2y+3xy^2)#

#=27x^3+81x^2y+81xy^2+27y^3#.

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Answer 2

Lincoln Anderson

To expand ( (3x + 3y)^3 ), we can use the binomial theorem or the distributive property. Let's use the distributive property:

[ (3x + 3y)^3 = (3x + 3y)(3x + 3y)(3x + 3y) ]

Expanding this expression, we multiply each term in the first parentheses by each term in the second parentheses, and then by each term in the third parentheses:

[ (3x + 3y)^3 = (3x)(3x)(3x) + (3x)(3x)(3y) + (3x)(3y)(3x) + (3x)(3y)(3y) + (3y)(3x)(3x) + (3y)(3x)(3y) + (3y)(3y)(3x) + (3y)(3y)(3y) ]

Simplifying each term, we get:

[ (3x)^3 + (3x)^2(3y) + (3x)(3y)^2 + (3y)^3 ]

[ = 27x^3 + 27x^2y + 27xy^2 + 27y^3 ]

Therefore, ( (3x + 3y)^3 ) expands to ( 27x^3 + 27x^2y + 27xy^2 + 27y^3 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.