You need to use the binomial formula, such that:
`(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k) y^k`
You need to replace c for x, d for y and 3 for n, such that:
`(c+d)^3 = 3C0 (c)^3+3C1 (c)^2*(d)^1+3C2 (c)^1*(d)^2+3C3(d)^3`
By definition, nC0 = nCn = 1, hence `3C0 = 3C3 = 1.`
By definition nC1 = nC(n-1) = n, hence `3C1 = 3C2 = 3.`
`(c+d)^3 = c^3 + 3c^2*d+ 3c*d^2+d^3`
Hence, expanding the complex number using binomial theorem yields the simplified result `(c+d)^3 = c^3 + 3c^2*d+ 3c*d^2+d^3` .
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