Respuesta :

apologiabiology
you would expand becausee it isn't equal to anything

(x+3)^3=(x+3)(x+3)(x+3)
we can use binomial theorem or pascal's triangle
anyway, pascal's triangle is more practicl for this one

so
for the 3rd power
[tex](a+b)^3=1a^3+3a^2b^1+3a^1b^2+1b^3[/tex]

a=x
b=3

[tex](x+3)^3=1x^3+3x^2(3^1)+3x^1(3)^2+1(3)^3[/tex]=
[tex]x^3+9x^2+27x+27[/tex]


garydesir1
Hi

(x + 3)³
(x+3)*(x+3)*(x+3)
(x+3)*(x²+6x+9)
Now you need to use the distributive property
(x)(x²) + (x)(6x) + (x)(9) +(3)(x²) + (3)(6x) + (3)(9)
Calculate them
x³ + 6x² + 9x + 3x² + 18x + 27
Group the common there
x³ + 6x² + 3x² + 9x + 18x + 27
x³ + 9x² + 27x + 27

I hope that's help !

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