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Question
Find the cube of: `"x"-1/2`
Sum
Solution
`("x"-1/2)^3`
`=("x")^3-(1/2)^3-3xx"x"xx1/2("x"-1/2)`
`="x"^3-1/8-(3"x")/2("x"-1/2)`
`="x"^3-1/8-(3"x"^2)/2+(3"x")/4`
`="x"^3-(3"x"^2)/2+(3"x")/4-1/8`
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Cubes of Binomials
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