Find the fourth proportion of $x^{2} - y^{2}, x^{2} - x y, x^{3} + y^{3}$ .

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Solution

Let the fourth proportion be $z$ . Then

$\frac{x^{2} - y^{2}}{x^{2} - x y} = \frac{x^{3} + y^{3}}{z}$

$\Rightarrow z = \frac{(x^{3} + y^{3}) x (x - y)}{x^{2} - y^{2}} = \frac{(x + y) (x^{2} - x y + y^{2}) x (x - y)}{(x + y) (x - y)}$

$z = x (x^{2} - x y + y^{2}) . . . . . . [∵ a^{2} - b^{2} = (a + b) (a - b) a^{3} + b^{3} = (a + b) (a^{2} + b^{2} = a b)]$

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Similar questions

[\frac{x^{3} + y^{3}}{{(x - y)}^{2} + 3 x y}] \div [\frac{{(x + y)}^{2} - 3 x y}{x^{3} - y^{3}}] \times \frac{x y}{x^{2} - y^{2}}

[\frac{x^{3} + y^{3}}{(x - y)^{2} + 3 x y}] + [\frac{(x + y)^{2} - 3 x y}{x^{3} - y^{3}}] \times \frac{x y}{x^{2} - y^{2}}

[\frac{x^{3} + y^{3}}{{(x - y)}^{2} + 3 x y}] \div [\frac{{(x + y)}^{2} - 3 x y}{x 3 - y^{3}}] \times \frac{x y}{x^{2} - y^{2}}

\frac{x^{2} + y^{2}}{x^{2} - y^{2}} = \frac{17}{8}

\frac{x^{3} + y^{3}}{x^{3} - y^{3}}

x^{3} + y^{3}

x^{2} - x y + y^{2}

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