Answer to Question #225880 in Algebra for Meerab

Question

expand by using binomial theorem: (x^2/2-2/x)^4

Answer 1

$\left( \cfrac {x^2} 2 - \cfrac 2 x \right) ^{ \large 4} = \left( \cfrac {x^2} 2 \right)^{ \large 4} - \large{4} \left(\cfrac {x^2} 2 \right)^{ \large 3}\cfrac 2 x + \large{6} \left( \cfrac {x^2} 2 \right)^{ \large 2}\left( \cfrac 2 x \right)^{ \large 2} - \large{4}\left( \cfrac {x^2} 2 \right) \left( \cfrac 2 x \right)^{ \large 3} +\\ +\left( \cfrac 2 x \right)^{ \large 4} = \cfrac {x^8} {16} - \large{4} \left(\cfrac {x^6} 8 \right) \cfrac 2 x + \large{6} \left( \cfrac {x^4} 4 \right)\left( \cfrac 4 {x^2} \right) - \large{4}\left( \cfrac {x^2} 2 \right) \left( \cfrac 8 {x^3} \right) + \cfrac {16} {x^4} = \\ = \cfrac {x^8} {16} - x^5 + \large{6} x^2 - \cfrac {16} x + \cfrac {16} {x^4}$