how do i solve: (c-d) to the power of 3?

Question

it has to do with the binomial theorem.. please help..

Answer 1

It's not an equation, so you can't "solve" it. You can write out the expression though.

The binomial theorem relates to rows of Pascal's triangle. Numbering the rows starting with 0, the jth entry of the ith row is i-choose-j. And the nth row also tells you the coefficients of (a+b)^n when it's expanded. The first few rows of the triangle are:

1
1 1
1 2 1
1 3 3 1

The 3rd row (again, starting the indexing at 0 instead of 1) is "1 3 3 1". So this means (c-d)^3, which is (c + (-d))^3, is:

1*[ c^3 (-d)^0 ] +3*[c^2 (-d)^1] + 3*[c^1 (-d)^2 ] + 1*[c^0 (-d)^3]

This simplifies to
c^3 - 3(c^2)d + 3cd^2 - d^3

Answer 2

for powers of 3 from the binomial formula,
the distribution of coefficients goes like
+1 -3 +3 -1
in decreasing powers of c and increasing powers of d.

Here,
(+1)*c^3 -3*(c^2)*(d) +3*c*d^2 -1*d^3
is the expansion

Answer 3

The binomial theorem states
(a + b)^n =
n
∑ (nCk)a^kb^(n - k)
k=0
In this case, n=3, a = c, and b = - d
Expanding the series you arrive at
(c + (-d))^3 = (3C0)c^3d^0 + (3C1)(c^2)(-d)^1 + (3C2)(c^1)(-d)^2 + (3C3)(c^1)(-d)^3 =
c^3 - 3c^2d + 3cd^2 - d^3