Question

factorise the following​

Answer 1

Answer:

see the attachment...

Hope it's useful....

Answer 2

$\huge\mathbb{\underline{\underline{QUESTION:-}}}$

Factorise:-

$\sf 1) a{}^{4}-(b+c){}^{4}\\ \sf 2) l{}^{2}-(m-n){}^{2}\\ \sf 3) x{}^{4}-2x{}^{2}y{}^{2}+y{}^{4}\\ \sf 4) 4(a+b){}^{2}-9(a-b){}^{2}$

$\huge \bf\red{\mid{ \overline {\underline{SOLUTION:-}}}\mid}$

$\sf 1 )a{}^{4}-(b+c){}^{4}\\ \sf Let\:(b+c)=x\\ \sf\implies a{}^{4}-x{}^{4}\\ \sf\implies (a{}^{2}){}^{2}-(x{}^{2}){}^{2}\\ \sf By \: formula\:of a{}^{2}-b{}^{2}=(a+b)(a-b)\\ \sf\implies( a{}^{2}+x{}^{2}) (a{}^{2}-x{}^{2})\\ \sf\implies( a{}^{2}+x{}^{2})(a+x)(a-x)\\ \sf Put\:the\: value\:of\:x\\ \sf\implies[a{}^{2}+(b+c){}^{2}][(a+b+c)(a-(b+c)\\ \sf\implies (b+c){}^{2}=b{}^{2}+2bc+c{}^{2}\\ \sf\implies [a{}^{2}+b{}^{2}+2bc+c{}^{2}][(a+b+c)(a-b-c)]$

$\sf 2) l{}^{2}-(m-n){}^{2}\\ \sf Let\: (m-n)=x\\ \sf\implies l{}^{2}-x{}^{2}\\ \sf\implies (l+x)(l-x)\\ \sf Put\: the\:value\:of\:x\\ \sf\implies [l+(m-n)][l-(m-n)]\\ \sf\implies(l+m-n)(l-m+n)$

$\sf 3) x{}^{4}-2x{}^{2}+y{}^{2}+y{}^{4}\\ \sf\implies (x{}^{2}-y{}^{2}){}^{2}\\ \sf By\:(a-b){}^{2}:-a{}^{2}-2ab+b{}^{2} \\ \sf\implies (x{}^{2}){}^{2}-2×x{}^{2}×y{}^{2}+(y{}^{2}){}^{2}\\ \sf\implies (x{}^{2}-y{}^{2}){}^{2}\\ \sf By\: a{}^{2}-b{}^{2}:(a+b)(a-b)\\ \sf\implies((x+y){}^{2}(x-y){}^{2})</p><p>$

$\sf (4) 4(a+b){}^{2}-9(a-b){}^{2}\\ \sf\implies</p><p>(2a+2b){}^{2}-(3a-3b){}^{2}\\ \sf\implies (2a+2b+3a-3b)(2a+2b-3a+3b)\\ \sf\implies (5a-b)(5b-a)$

$\sf\large{\blue{1=[a{}^{2}+b{}^{2}+2bc+c{}^{2}][(a+b+c)(a-b-c)] }}$

$\sf\large{\pink{2 =(l+m-n)(l-m+n)}}$

$\sf\large{\blue{3=((x+y){}^{2}(x-y){}^{2})}}$

$\sf\large{\pink{4=(5a-b)(5b-a)}}$.