化简 (\sqrt[3]{2}a-sqrt(2)b)^2

解答

化简

(32 a - 2 b)^{2}

3 2^{2} a^{2} - 6 2^{5} \cdot 2 ab + 2 b^{2}

求解步骤

(32 a - 2 b)^{2}

使用完全平方公式:

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

(32 a - 2 b)^{2} = (32 a)^{2} - 232 a 2 b + (2 b)^{2}

= (32 a)^{2} - 232 a 2 b + (2 b)^{2}

化简

(32 a)^{2} - 232 a 2 b + (2 b)^{2} : 2^{\frac{2}{3}} a^{2} - 2 \cdot 2^{\frac{5}{6}} ab + 2 b^{2}

= 2^{\frac{2}{3}} a^{2} - 2 \cdot 2^{\frac{5}{6}} ab + 2 b^{2}

= 2^{\frac{2}{3}} a^{2} - 2^{\frac{5}{6}} \cdot 2 ab + 2 b^{2}

2^{\frac{2}{3}} = 3 2^{2}

= 3 2^{2} a^{2} - 6 2^{5} \cdot 2 ab + 2 b^{2}

2^{\frac{5}{6}} = 6 2^{5}

= 3 2^{2} a^{2} - 6 2^{5} \cdot 2 ab + 2 b^{2}

s im pl i f y (2 + x)^{2}

e x p an d (1 + cos (t))^{2}

e x p an d \frac{x ^{4} + 6 x - 1}{( x + 1 ) ( x ^{2} - 2 x + 4 )}

e x p an d (a + b + c)^{5}

s im pl i f y (cos (x) + i sin (x))^{4}