How do you use the binomial theorem to expand and simplify the expression (c+d)^3?

1 Answer
Jan 3, 2018

${c}^{3} + 3 {c}^{2} d + 3 c {d}^{2} + {d}^{3}$

Explanation:

1 ${\left(c + d\right)}^{0}$
1 1 ${\left(c + d\right)}^{1}$
1 2 1 ${\left(c + d\right)}^{2}$
1 3 3 1 ${\left(c + d\right)}^{3}$

${\left(c + d\right)}^{0} = 1$
${\left(c + d\right)}^{1} = 1 \left(c\right) + 1 \left(d\right)$
${\left(c + d\right)}^{2} = 1 \left({c}^{2}\right) \left({d}^{0}\right) + 2 \left({c}^{1}\right) \left({d}^{1}\right) + 1 \left({c}^{0}\right) \left({d}^{2}\right)$
${\left(c + d\right)}^{3} = 1 \left({c}^{3}\right) \left({d}^{0}\right) + 3 \left({c}^{2}\right) \left({d}^{1}\right) + 3 \left({c}^{1}\right) \left({d}^{2}\right) + 1 \left({c}^{0}\right) \left({d}^{3}\right)$

${\left(c + d\right)}^{0} = 1$
${\left(c + d\right)}^{1} = c + d$
${\left(c + d\right)}^{2} = 1 \left({c}^{2}\right) \cancel{\left({d}^{0}\right)} + 2 \left({c}^{1}\right) \left({d}^{1}\right) + 1 \cancel{\left({c}^{0}\right)} \left({d}^{2}\right)$
${\left(c + d\right)}^{3} = 1 \left({c}^{3}\right) \cancel{\left({d}^{0}\right)} + 3 \left({c}^{2}\right) \left({d}^{1}\right) + 3 \left({c}^{1}\right) \left({d}^{2}\right) + 1 \cancel{\left({c}^{0}\right)} \left({d}^{3}\right)$

${\left(c + d\right)}^{0} = 1$
${\left(c + d\right)}^{1} = c + d$
${\left(c + d\right)}^{2} = {c}^{2} + 2 c d + {d}^{2}$
${\left(c + d\right)}^{3} = {c}^{3} + 3 {c}^{2} d + 3 c {d}^{2} + {d}^{3}$