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2015-02-02T15:03:37+05:30

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Let the third proportional be y.

 \frac{a-b}{a^2-b^2} = \frac{a^2-b^2}{y} \\ \\ \frac{a-b}{(a-b)(a+b)} = \frac{a^2-b^2}{y} \\ \\ \frac{1}{a+b} = \frac{a^2-b^2}{y} \\ \\y=(a+b)(a^2-b^2) \\ \\y=a^3-ab^2+a^2b-b^3\\ \\Third\ proportional\ is\ \boxed{a^3-ab^2+a^2b-b^3}
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2015-02-02T16:52:52+05:30
Let the third propotional be p
ATQ,
a-b:a²-b²::a²-b²:a
product of extreme terms = product of mean terms
p*(a-b)=(a²-b²)(a²-b²)
p*(a-b)=(a+b)(a-b)(a²-b²)
p=(a+b)(a-b)(a²+b²)/(a-b)
p=(a+b)(a²-b²)
p=a(a²-b²)+b(a²-b²)
p=a³-ab²+ba²-b³
Therefore the third proportion is a³-ab²+ba²-b³.

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