The number of pairs (x,y) satisfying the equations 1x3+1y3=35,1x2−1xy+1y2=7are

∵a3+b3=(a+b)(a2−ab+b2) 35 =(1x+1y)(1x2−1xy+1y2)1x+1y=5 putting 1y=5-1x in one of the given two equations, we get 2 Paris of solutions(12,13) and (13,12)

The number of pairs (x,y) satisfying the equations 1x3+1y3=35,1x2−1xy+1y2=7are

∵a3+b3=(a+b)(a2−ab+b2) 35 =(1x+1y)(1x2−1xy+1y2)1x+1y=5 putting 1y=5-1x in one of the given two equations, we get 2 Paris of solutions(12,13) and (13,12)

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The number of pairs (x,y) satisfying the equations $\frac{1}{x^{3}} + \frac{1}{y^{3}} = 35, \frac{1}{x^{2}} - \frac{1}{x y} + \frac{1}{y^{2}} = 7$ are

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Detailed Solution

$\begin{array}{l} ∵ a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2}) \\ 35 = (\frac{1}{x} + \frac{1}{y}) (\frac{1}{x^{2}} - \frac{1}{x y} + \frac{1}{y^{2}}) \end{array}$
$\frac{1}{x} + \frac{1}{y}$ =5 putting $\frac{1}{y}$ =5- $\frac{1}{x}$ in one of the given two equations, we get 2 Paris of solutions $(\frac{1}{2}, \frac{1}{3})$ and $(\frac{1}{3}, \frac{1}{2})$