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(x_0)^3+(y_0)^3=(z_0)^3. if 2p=x_0+y_0 and 2q=x_0-y_0, x_0=? and y_0=?
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Adding \(x_0+y_0=2p \text{ and } x_0-y_0=2q\) gives \(2x_0=2p+2q \implies x_0=p+q\). Substitute that in either equation to get \(y_0=2p-x_0=2p-p-q \implies y_0=p-q.\)
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