Question

2a6-128a3/(2a2+8a+32)*(a4-4a3) упростить пж

Answer 1

$frac{2 {a}^{6} - 128 {a}^{3} }{(2 {a}^{2} + 8 a + 32)( {a}^{4} - 4 {a}^{3}) } = frac{ 2{a}^{3}({a}^{3} - 64 {a}^{3} ) }{2( {a}^{2} + 4a + 16) times {a}^{3} (a - 4)} = frac{2(a - 4)( {a}^{2} + 4a + 16)}{2( {a}^{2} + 4a + 16)(a - 4)} = 1$