Question

How do you solve this problem? (x+3) cubed

Answer 1

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
$(a+b)^3=1a^3+3a^2b^1+3a^1b^2+1b^3$

a=x
b=3

$(x+3)^3=1x^3+3x^2(3^1)+3x^1(3)^2+1(3)^3$=
$x^3+9x^2+27x+27$

Answer 2

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 !