can someone please help me with this?

Question

(x+c)^3rd power.... how would that work out?

Answer 1

(x + c)^3
= (x + c)(x + c)(x +c)
= (x + c)(x^2 + cx + cx) + c^2)
= (x + c)(x^2 + 2cx + c^2)
= x^3 + 2c(x^2) + (c^2)x + c(x^2) + 2(c^2)x + c^3
= x^3 + 3c(x^2) + 3(c^2)x + c^3

You could also use the Binomial theorem or Pascal's triangle to figure out the terms.

Good luck! =)

Answer 2

First thing is to expand it, ike so:

(x+c)^3 = (x+c) * (x+c) * (x+c)

then you distribute the multiplication

(x+c)*(x+c) = x*(x+c) + c*(x+c) = x^2+xc+xc+c^2 = x^2+2cx+c^2

and

(x+c)*(x^2+2cx+c^2) = x^3+2cx^2+xc^2 + cx^2+2xc^2+c^3

and regrouping

x^3 + 3cx^2 + 3xc^2 + c^3

Answer 3

(x+c)^3= (x+c)(x+c)(x+c) so if you should take one part at a time and use the foil method.
(x+c)(x+c) will come out to be x^2+xc+xc+c^2 or x^2+2xc+c^2
then take (x^2+2xc+c^2)(x+c)
so x^3+cx^2+2cx^2+2xc^2+xc^2+c^3 or x^3+3cx^2+3xc^2+c^3

Answer 4

u can use binomial expansion(a+b)^n=
a^n+(n/1!)a^(n-1)b+[(n(n-1)/2!]a^(n-2) b ^2+[(n(n-1)(n-2))/3!]a ^(n-3)b^3..+b^n

(a+b)³ = a³ +(3/1!) a²b+[3(2)/2!]ab² +[(3(2)(1))/3!] b³

(a+b)³ = a³ + 3 a²b +3ab² +b³, (x+c)³ = x³ + 3x²c+3xc² +c³

hope you will understand the steps

Answer 5

(x+c)^3 =

(x+c)(x+c)(x+c) =

(x+c)(x^2 + c^2 + 2xc) =

x^3 + xc^2 + 2cx^2 + cx^2 + c^3 + 2xc^2 =

x^3 + c^3 + 3cx^2 + 3xc^2

I hope this helps! =)

Answer 6

Not sure of your question; I will expand:

= [(x+c)(x+c)(x+c)]
= (x^2 + 2xc + c^2)(x+c)
= x^3 + 2(x^2)c + (c^2)x + (x^2)c + 2x(c^2) + c^3
= x^3 + 3(x^2)c + 3x(c^2) + c^3

Hope this helps.

Answer 7

x^3 +3 x^2c +3xy^2+c^3

Answer 8

(x+c)^3 = 0

x = -c (triple root)