expand by using binomial theorem: (x^2/2-2/x)^4
(x22−2x)4=(x22)4−4(x22)32x+6(x22)2(2x)2−4(x22)(2x)3++(2x)4=x816−4(x68)2x+6(x44)(4x2)−4(x22)(8x3)+16x4==x816−x5+6x2−16x+16x4\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}(2x2−x2)4=(2x2)4−4(2x2)3x2+6(2x2)2(x2)2−4(2x2)(x2)3++(x2)4=16x8−4(8x6)x2+6(4x4)(x24)−4(2x2)(x38)+x416==16x8−x5+6x2−x16+x416
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