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

[♥anna♥]

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

[Geezah]

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:11 11 2 11 3 3 1The 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

[Not Eddie Money]

for powers of 3 from the binomial formula, the distribution of coefficients goes like+1 -3 +3 -1in decreasing powers of c and increasing powers of d.Here,(+1)*c^3 -3*(c^2)*(d) +3*c*d^2 -1*d^3is the expansion

[Helmut]

The binomial theorem states(a + b)^n = n∑ (nCk)a^kb^(n - k)k=0In this case, n=3, a = c, and b = - dExpanding 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