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Question Number 204590 by sphelele last updated on 22/Feb/24

Answered by A5T last updated on 22/Feb/24

$$a)=\frac{2^{n+2+2n+2}}{2^{3n-3}}=\frac{2^{3n+4}}{2^{3n+4}{2}^{-7}}=128$$$$\text{Missing left or extra right}$$$$c)=\frac{3^{m+1}(3^3-6)}{7\times 3\times 3^{m+1}}=1$$$$\left(d\right)=\frac{\sqrt{p-q}\left(\sqrt{p+q}\right){(p+q)}^{\frac{5}{2}}}{\sqrt{p-q}}={(p+q)}^3=p^3+q^3+3pq(p+q)$$

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