Solve the following equation.
$x^{3} - y^{3} = 127, x^{2} y - x y^{2} = 42$ .

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Solution

Putting $y = m x$ , the given equations can be written as $x^{3} (1 - m^{3}) = 127$ . $(1)$
and $x^{3} (m - m^{2}) = 42$ $(2)$
Dividing, we get
$\frac{1 - m^{3}}{m - m^{2}} = \frac{127}{42}$ or $\frac{1 + m + m^{2}}{m} = \frac{127}{42}$
or $42 m^{2} - 85 m + 42 = 0$
or $42 m^{2} - 49 m - 36 m + 42 = 0$
or $(6 m - 7) (7 m - 6) = 0$ .
$∴ m = 7 / 6$ or $6 / 7$ .
We first take $m = 7 / 6$ and we get from $(1)$
$x^{3} (1 - 343 / 216) = 127$ or $x = - 6$
and $y = m x = \frac{7}{6} (- 6) = - 7$ .
Similarly taking $m = 6 / 7$ , we shall get $x = 7$ , $y = 6$ .
Hence solutions are
$x = 7, y = 6$ or $x = - 6, y = - 7$ .
Alternative method:
Multiply the second equation by $(3)$ and subtract it from first and we get
$(x - y)^{3} = 1$ , $∴ x - y = 1$ or $x = y + 1$
Putting in $2$ nd equation, i.e., $x y (x - y) = 42$ , we get
$y (y + 1) \cdot 1 = 42$ or $y^{2} + y - 42 = 0$
$∴ y = - 7, 6$ and hence $x = - 6, 7$
$∴ (7, 6); (- 6, - 7)$ .

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