问题
Simplify $\left(7 b^{3}\right)^{2} \cdot\left(4 b^{2}\right)^{-3}$, given that bis non-zero. 标准答案

Answer

Expert–verified

Hide Steps

Answer

Since $b^0 = 1$, the final answer is $\boxed{\frac{49}{64}}$

Steps

Step 1 ：Apply the power of a power rule: $(a^m)^n = a^{mn}$

Step 2 ：$\left(7 b^{3}\right)^{2} = 7^2 b^{3 \cdot 2} = 49 b^6$

Step 3 ：$\left(4 b^{2}\right)^{-3} = 4^{-3} b^{2 \cdot -3} = \frac{1}{64} b^{-6}$

Step 4 ：Combine the terms: $49 b^6 \cdot \frac{1}{64} b^{-6}$

Step 5 ：Apply the product of powers rule: $a^m \cdot a^n = a^{m+n}$

Step 6 ：$\frac{49}{64} b^{6-6} = \frac{49}{64} b^0$

Step 7 ：Since $b^0 = 1$, the final answer is $\boxed{\frac{49}{64}}$