Squares with a difference
p 2 = q 2 + k , p = q
Then ( p + q ) 2 ( p − q ) 2 p 2 − q 2 = ?
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2 solutions
You should change one of the terms in denominator to (p-q) in your solution.
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URghh so many typos. Thanks!
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Haha. And in the numerator one of them should be (p+q) so that it can cancel out with one of the similar terms in denominator.
I factorized, got to the answer and then clicked the wrong one!
I did the same way!
You should mention that p=!q
( ( p + q ) ( p − q ) ) 2 ( p + q ) ( p − q )
= ( p + q ) ( p − q ) 1
= ( p 2 − q 2 ) 1
= k 1
Sorry, I have edited the typo, the answer should now be k 1 , Sorry for the inconvenience
p 2 = q 2 + k ⟹ p 2 − q 2 = k ⟹ ( p + q ) ( p − q ) = k
\[\begin{align}&= \dfrac{p^2-q^2}{(p+q)^2 \cdot (p-q)^2} \\ &= \dfrac{(p+q)(p-q)}{\big((p+q) \cdot (p-q)\big)^2} \\ &= \dfrac{{\color{Red}{k}}}{({\color{Red}{k}})^2} \\ &= \dfrac{1}{{\color{Red}{k}}}
\end{align}\]