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

Answer

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Answer

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

Steps

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

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

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

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 $\frac{49}{64}$