Find the compounded ratio of the following:
ii) $(a^{2} - b^{2}) : (a^{2} + b^{2}) and (a^{4} - b^{4}) : (a + b)^{4}$

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Solution

Given: $(a^{2} - b^{2}) : (a^{2} + b^{2}) and (a^{4} - b^{4}) : (a + b)^{4}$

$(a^{2} - b^{2}) : (a^{2} + b^{2}) = \frac{a^{2} - b^{2}}{a^{2} + b^{2}} = \frac{(a - b) (a + b)}{a^{2} + b^{2}}$

$(a^{4} - b^{4}) : (a + b)^{4} = \frac{(a^{4} - b^{4})}{(a + b)^{4}}$

$= \frac{(a^{2} - b^{2}) (a^{2} + b^{2})}{(a + b)^{4}} = \frac{(a + b) (a - b) (a^{2} + b^{2})}{(a + b)^{4}}$

So, the compounded ratio of

$\frac{a^{2} - b^{2}}{a^{2} + b^{2}} and \frac{(a^{4} - b^{4})}{(a + b)^{4}}$ ,

$= \frac{a^{2} - b^{2}}{a^{2} + b^{2}} \times \frac{(a^{4} - b^{4})}{(a + b)^{4}}$

$= \frac{(a - b) (a + b)}{a^{2} + b^{2}} \times \frac{(a + b) (a - b) (a^{2} + b^{2})}{(a + b)^{4}}$

$= \frac{(a - b)^{2} (a + b)^{2} (a^{2} + b^{2})}{(a^{2} + b^{2}) (a + b)^{4}}$

$= \frac{(a - b)^{2}}{(a + b)^{2}}$

Hence, the compounded ratio of $(a^{2} - b^{2}) : (a^{2} + b^{2}) and (a^{4} - b^{4}) : (a + b)^{4}$ is $(a - b)^{2} : (a + b)^{2}$ .

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Question 92 (xviii)

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

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

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

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}

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a^{4} + a^{2} b^{2} + b^{4}

a^{2} + b^{2} + a b

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