What is (2/(b^2))^3*((b^3)/4)^2*(1/(b^2))^2 ?

The solution to (2/(b^2))^3*((b^3)/4)^2*(1/(b^2))^2 is 1/(2b^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (2/(b^2))^3((b^3)/4)^2(1/(b^2))^2

simplify

(\frac{2}{b ^{2}})^{3} \cdot (\frac{b ^{3}}{4})^{2} \cdot (\frac{1}{b ^{2}})^{2}

\frac{1}{2 b ^{4}}

Solution steps

(\frac{2}{b ^{2}})^{3} (\frac{b ^{3}}{4})^{2} (\frac{1}{b ^{2}})^{2}

Simplify

(\frac{2}{b ^{2}})^{3} : \frac{8}{b ^{6}}

= \frac{8}{b ^{6}} (\frac{b ^{3}}{4})^{2} (\frac{1}{b ^{2}})^{2}

Simplify

(\frac{b ^{3}}{4})^{2} : \frac{b ^{6}}{16}

= \frac{8}{b ^{6}} \cdot \frac{b ^{6}}{16} (\frac{1}{b ^{2}})^{2}

Simplify

(\frac{1}{b ^{2}})^{2} : \frac{1}{b ^{4}}

= \frac{8}{b ^{6}} \cdot \frac{b ^{6}}{16} \cdot \frac{1}{b ^{4}}

Cross-cancel common factor:

b^{6}

= \frac{8}{16} \cdot \frac{1}{b ^{4}}

Cancel the numbers:

\frac{8}{16} = \frac{1}{2}

= \frac{1}{2} \cdot \frac{1}{b ^{4}}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{1 \cdot 1}{2 b ^{4}}

Multiply the numbers:

1 \cdot 1 = 1

= \frac{1}{2 b ^{4}}

(3 + 2 i) + [(4 - 5 i) - (5 + i)]

\frac{a + b}{a ^{2} + ab} - \frac{b - a}{ab + b ^{2}}

8 u^{4} + u

(4 x + 3 y) (6 x - 5 y)

\frac{1}{x - 1} - \frac{1}{x}

What is (2/(b^2))^3*((b^3)/4)^2*(1/(b^2))^2 ?
The solution to (2/(b^2))^3*((b^3)/4)^2*(1/(b^2))^2 is 1/(2b^4)