Find the binomial expansion of (-8+4x)^(2/3) up to and including the term in x^2.

The formula booklet is useful for this type of question, but only once we've rearranged the expression into the correct form. The formula booklet expression is: (1 + z)^n = 1 + nz + 0.5n*(n-1)z^2 + .... To start, we need to factorise the expression in the question so that ultimately the constant in the brackets is equal to '1'. Factorising: [-8+4x]^(2/3) = [(-8)(1 - x/2)]^(2/3). We then use the laws of indices to move the '-8' outside the expression: [(-8)(1 - x/2)]^(2/3) = (-8)^(2/3) * (1 - x/2)^(2/3) = 4 * (1 - x/2)^(2/3). We now have the expression in the correct form, with z = -x/2 and n = 2/3. Substituting these values into the formula booklet's expression (and remembering to multiply by the 4) we get: 4 * [(1 - x/2)^(2/3)] = 4 * [1 + (2/3)(-x/2) + 0.5*(2/3)(2/3-1)(-x/2)^2] = 4 * [1 - x/3 - (x^2)/36] = 4 - 4x/3 - (x^2)/9.This is the final result: (-8+4x)^(2/3) = 4 - 4x/3 - (x^2)/9