Factorise : $a^{4} - (a^{2} - 3 b^{2})^{2}$

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Solution

We have, $a^{4} - (a^{2} - 3 b^{2})^{2}$
$= (a^{2})^{2} - (a^{2} - 3 b^{2})^{2}$
$= [a^{2} + (a^{2} - 3 b^{2})] [a^{2} - (a^{2} - 3 b^{2})] [Using the identity (x^{2} - y^{2}) = (x - y) (x + y)]$
$= (a^{2} + a^{2} - 3 b^{2}) (a^{2} - a^{2} + 3 b^{2})$
$= (2 a^{2} - 3 b^{2}) (3 b^{2})$
$= 3 b^{2} (2 a^{2} - 3 b^{2})$
$= 3 b^{2} [(\sqrt{2} a)^{2} - (\sqrt{3} b)^{2}] [Using above identity]$
$= 3 b^{2} (\sqrt{2} a - \sqrt{3} b) (\sqrt{2} a + \sqrt{3} b)$

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Question 92 (xviii)

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

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a^{4} - b^{4}

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Factorise : a4−(a2−3b2)2

We have, a4−(a2−3b2)2=(a2)2−(a2−3b2)2=[a2+(a2−3b2)][a2−(a2−3b2)] [Using the identity(x2−y2)=(x−y)(x+y)]=(a2+a2−3b2)(a2−a2+3b2)=(2a2−3b2)(3b2)=3b2(2a2−3b2)=3b2[(√2a)2−(√3b)2] [Using above identity]=3b2(√2a−√3b)(√2a+√3b)

Factorise : $a^{4} - (a^{2} - 3 b^{2})^{2}$