Find the compound ratio of $a - b : a + b, (a + b)^{2} : a^{2} + b^{2}$ and $a^{4} - b^{4} : (a^{2} - b^{2})^{2}$ .

1 : 1

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1 : 2

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2 : 2

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2 : 1

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Solution

The correct option is A $1 : 1$
Required compound ratio $= [(a - b) . (a + b)^{2} . (a^{4} - b^{4})] : [(a + b) . (a^{2} + b^{2}) . (a^{2} - b^{2})^{2}]$
$= \frac{(a - b) (a + b)^{2} (a^{2} + b^{2}) (a^{2} - b^{2})}{(a + b) (a^{2} + b^{2}) (a^{2} - b^{2}) (a + b) (a - b)}$ $= 1 : 1$

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Similar questions

Question 92 (xviii)

Factorise the following using the identity $a^{2} - b^{2} = (a + b) (a - b) .$

$a^{4} - (a - b)^{4}$

If X = 1 - a^{2} and Y = 1 - b^{2}, then what is X^{2} + Y^{2} ?

a^{4} b - a b^{4}, a^{4} b^{2} - a^{2} b^{4}

a^{2} b^{2} (a^{4} - b^{4})

(a + b) : (a - b) = 1 : 5

(a^{2} - b^{2}) : (a^{2} + b^{2}) .

\frac{a^{2} + a b + b^{2}}{b^{2} + b c + c^{2}} = \frac{a}{c}

\frac{a^{4} + a^{2} b^{2} + b^{4}}{b^{4} + b^{2} c^{2} + c^{4}} = \frac{a^{2}}{c^{2}}

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