展开 (2y-y'(t))^3

解答

展开

(2 y - y^{'} (t))^{3}

8 y^{3} - 12 y^{2} y' (t) + 6 y (y' (t))^{2} - (y' (t))^{3}

求解步骤

(2 y - y^{'} (t))^{3}

使用完全立方公式:

(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}

a = 2 y, b = y^{'} (t)

= (2 y)^{3} - 3 (2 y)^{2} y^{^{'}} (t) + 3 \cdot 2 y (y^{^{'}} (t))^{2} - (y^{^{'}} (t))^{3}

化简

(2 y)^{3} - 3 (2 y)^{2} y^{'} (t) + 3 \cdot 2 y (y^{'} (t))^{2} - (y^{'} (t))^{3} : 8 y^{3} - 12 y^{2} y^{'} (t) + 6 y (y^{'} (t))^{2} - (y^{'} (t))^{3}

= 8 y^{3} - 12 y^{2} y^{'} (t) + 6 y (y^{'} (t))^{2} - (y^{'} (t))^{3}

e x p an d \frac{2 ( - 3 x ^{2} + 7 )}{( x ^{2} + 7 ) ^{3}}

e x p an d \frac{3}{( 4 + n ) ( 5 + n )}

e x p an d \frac{( 2 x - 3 ) ( x + 3 )}{2}

e x p an d \frac{s ^{4} + 26 s ^{2} + 64}{s ( s ^{2} + 4 ) ( s ^{2} + 16 )}

e x p an d (y^{3})^{2}